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Question Number 111859 by ajfour last updated on 05/Sep/20

I=∫(dx/((x^2 +2x+3)(√(x^2 +x+3)))) = ? my try..

$${I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}\:=\:? \\ $$$${my}\:{try}.. \\ $$

Commented by Sarah85 last updated on 05/Sep/20

my try: (1.) u=ln (2x+1+2(√(x^2 +x+3))) x=((e^(2u) −2e^u −11)/(4e^u )) dx=(√(x^2 +x+3))du 16∫(e^(2u) /(e^(4u) +4e^(3u) +14e^(2u) −44e^u +121))du (2.) v=e^u u=ln v du=(dv/e^u ) 16∫(v/(v^4 +4v^3 +14v^2 −44v+121))dv of course now we must decompose and it′s theoretically easy but hard to write out... I don′t think there′s an alternative without a 4^(th) −degree polynome...

$$\mathrm{my}\:\mathrm{try}: \\ $$$$\left(\mathrm{1}.\right)\:{u}=\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:{x}=\frac{\mathrm{e}^{\mathrm{2}{u}} −\mathrm{2e}^{{u}} −\mathrm{11}}{\mathrm{4e}^{{u}} } \\ $$$$\:\:\:\:\:\:\:\:\:{dx}=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}{du} \\ $$$$\mathrm{16}\int\frac{\mathrm{e}^{\mathrm{2}{u}} }{\mathrm{e}^{\mathrm{4}{u}} +\mathrm{4e}^{\mathrm{3}{u}} +\mathrm{14e}^{\mathrm{2}{u}} −\mathrm{44e}^{{u}} +\mathrm{121}}{du} \\ $$$$\left(\mathrm{2}.\right)\:{v}=\mathrm{e}^{{u}} \\ $$$$\:\:\:\:\:\:\:\:\:{u}=\mathrm{ln}\:{v} \\ $$$$\:\:\:\:\:\:\:\:\:{du}=\frac{{dv}}{\mathrm{e}^{{u}} } \\ $$$$\mathrm{16}\int\frac{{v}}{{v}^{\mathrm{4}} +\mathrm{4}{v}^{\mathrm{3}} +\mathrm{14}{v}^{\mathrm{2}} −\mathrm{44}{v}+\mathrm{121}}{dv} \\ $$$$\mathrm{of}\:\mathrm{course}\:\mathrm{now}\:\mathrm{we}\:\mathrm{must}\:\mathrm{decompose}\:\mathrm{and}\:\mathrm{it}'\mathrm{s} \\ $$$$\mathrm{theoretically}\:\mathrm{easy}\:\mathrm{but}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{write}\:\mathrm{out}... \\ $$$$\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{alternative}\:\mathrm{without} \\ $$$$\mathrm{a}\:\mathrm{4}^{\mathrm{th}} −\mathrm{degree}\:\mathrm{polynome}... \\ $$

Commented by MJS_new last updated on 05/Sep/20

good. I would add a 3^(rd) step w=v+1 → dv=dw 16∫((w−1)/(w^4 +8w^2 −64w+176))dw the square factors are w^2 −2(√(2+2(√3)))w+4(2+(√3)−2(√(−1+(√3)))) and w^2 +2(√(2+2(√3)))w+4(2+(√3)+2(√(−1+(√3)))) ⇒ 16∫((w−1)/(w^4 +8w^2 −64w+176))dw= =((√3)/6)∫(((√(−1+(√3)))w−4+2(√(2+2(√3))))/(w^2 −2(√(2+2(√3)))w+4(2+(√3)−2(√(−1+(√3))))))dw− −((√3)/6)∫(((√(−1+(√3)))w+4+2(√(2+2(√3))))/(w^2 +2(√(2+2(√3)))w+4(2+(√3)+2(√(−1+(√3))))))dw and we can solve these using the formula for ∫((ax+b)/(x^2 +cx+d))dx if we must; i.e. if our survival depends on it

$$\mathrm{good}.\:\mathrm{I}\:\mathrm{would}\:\mathrm{add}\:\mathrm{a}\:\mathrm{3}^{\mathrm{rd}} \:\mathrm{step} \\ $$$${w}={v}+\mathrm{1}\:\rightarrow\:{dv}={dw} \\ $$$$\mathrm{16}\int\frac{{w}−\mathrm{1}}{{w}^{\mathrm{4}} +\mathrm{8}{w}^{\mathrm{2}} −\mathrm{64}{w}+\mathrm{176}}{dw} \\ $$$$\mathrm{the}\:\mathrm{square}\:\mathrm{factors}\:\mathrm{are} \\ $$$${w}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}{w}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{−\mathrm{1}+\sqrt{\mathrm{3}}}\right) \\ $$$$\mathrm{and} \\ $$$${w}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}{w}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{−\mathrm{1}+\sqrt{\mathrm{3}}}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{16}\int\frac{{w}−\mathrm{1}}{{w}^{\mathrm{4}} +\mathrm{8}{w}^{\mathrm{2}} −\mathrm{64}{w}+\mathrm{176}}{dw}= \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\int\frac{\sqrt{−\mathrm{1}+\sqrt{\mathrm{3}}}{w}−\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}}{{w}^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}{w}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{2}\sqrt{−\mathrm{1}+\sqrt{\mathrm{3}}}\right)}{dw}− \\ $$$$−\frac{\sqrt{\mathrm{3}}}{\mathrm{6}}\int\frac{\sqrt{−\mathrm{1}+\sqrt{\mathrm{3}}}{w}+\mathrm{4}+\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}}{{w}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}{w}+\mathrm{4}\left(\mathrm{2}+\sqrt{\mathrm{3}}+\mathrm{2}\sqrt{−\mathrm{1}+\sqrt{\mathrm{3}}}\right)}{dw} \\ $$$$\mathrm{and}\:\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{these}\:\mathrm{using}\:\mathrm{the}\:\mathrm{formula}\:\mathrm{for} \\ $$$$\int\frac{{ax}+{b}}{{x}^{\mathrm{2}} +{cx}+{d}}{dx} \\ $$$$\mathrm{if}\:\mathrm{we}\:\mathrm{must};\:\mathrm{i}.\mathrm{e}.\:\mathrm{if}\:\mathrm{our}\:\mathrm{survival}\:\mathrm{depends}\:\mathrm{on}\:\mathrm{it} \\ $$

Commented by mathdave last updated on 06/Sep/20

u can only survive a way out only by using ferrari trick or idea to decompose d degree 4 to difference of a two square

$${u}\:{can}\:{only}\:{survive}\:{a}\:{way}\:{out}\:{only}\:{by} \\ $$$${using}\:{ferrari}\:{trick}\:{or}\:{idea}\:{to} \\ $$$${decompose}\:{d}\:{degree}\:\mathrm{4}\:{to}\:{difference}\:{of}\:{a} \\ $$$${two}\:{square}\: \\ $$

Commented by mathdave last updated on 06/Sep/20

should i help u break this 4 degree of this equation

$${should}\:{i}\:{help}\:{u}\:{break}\:{this}\:\mathrm{4}\:{degree}\:{of} \\ $$$${this}\:{equation} \\ $$

Commented by mathdave last updated on 06/Sep/20

who told u dat there is no alternative without degree 4 polynomial,if u wanna kwn check my working to see dat

$${who}\:{told}\:{u}\:{dat}\:{there}\:{is}\:{no}\:{alternative} \\ $$$${without}\:{degree}\:\mathrm{4}\:{polynomial},{if}\:{u} \\ $$$${wanna}\:{kwn}\:{check}\:{my}\:{working}\:{to} \\ $$$${see}\:{dat} \\ $$

Commented by MJS_new last updated on 06/Sep/20

no help needed, thank you very much. as you can see (if you take a closer look) I already decomposed the 4^(th) degree. all that′s left is inserting the constants in the well known formula. no need to re−invent the wheel.

$$\mathrm{no}\:\mathrm{help}\:\mathrm{needed},\:\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}. \\ $$$$\mathrm{as}\:\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\left(\mathrm{if}\:\mathrm{you}\:\mathrm{take}\:\mathrm{a}\:\mathrm{closer}\:\mathrm{look}\right)\:\mathrm{I} \\ $$$$\mathrm{already}\:\mathrm{decomposed}\:\mathrm{the}\:\mathrm{4}^{\mathrm{th}} \:\mathrm{degree}.\:\mathrm{all}\:\mathrm{that}'\mathrm{s} \\ $$$$\mathrm{left}\:\mathrm{is}\:\mathrm{inserting}\:\mathrm{the}\:\mathrm{constants}\:\mathrm{in}\:\mathrm{the}\:\mathrm{well} \\ $$$$\mathrm{known}\:\mathrm{formula}.\:\mathrm{no}\:\mathrm{need}\:\mathrm{to}\:\mathrm{re}−\mathrm{invent}\:\mathrm{the} \\ $$$$\mathrm{wheel}. \\ $$

Commented by mathdave last updated on 06/Sep/20

anyhow or watever

$${anyhow}\:{or}\:{watever} \\ $$

Answered by mathdave last updated on 06/Sep/20

solution to let I=∫(dx/((x^2 +2x+3)(√(x^2 +x+3)))) put x=(((√3)−(√3)t)/(1+t)) and dx=−((2(√3))/((1+t)^2 ))dt I=∫((−2(√3)dt)/((((3(1−t)^2 )/((1+t)^2 ))+2(√3)(((1−t))/((1+t)))+3)(√(((3(1−t)^2 )/((1+t)^2 ))+(√3)(((1−t)/(1+t)))+3)))) by simplify we have that I=∫((−2(√3)(1+t)dt)/((6−2(√3))t^2 +(6+2(√3)))(√((6−(√3))t^2 +(6+(√3)))))) let a=6−2(√3),b=6+2(√3),c=6−(√3),d=6+(√3) ∫((−2(√3)(1+t)dt)/((at^2 +b)(√(ct^2 +d))))=−2(√3)∫(dt/((at^2 +b)(√(ct^2 +d))))−2(√3)∫(t/((at^2 +b)(√(ct^2 +d))))dt Let A=∫(dt/((at^2 +b)(√(ct^2 +d)))) let y=(t/(√(ct^2 +d))) ,t^2 =((y^2 d)/(1−y^2 c)),dt=((√d)/((1−y^2 c)^(3/2) ))dy A=∫((y(√d))/(((1−y^2 c)^(3/2) )/([a(((y^2 d)/(1−y^2 c)))+b](√((y^2 d)/(1−y^2 c))))))dy= A=∫(dy/((ay^2 d+b−y^2 bc)(1−y^2 c))) A=∫(((bc−ad)dy)/(ad[(bc−ad)y^2 −b]))−∫((cdy)/(ad(cy^2 −1))) A=((ad−bc)/(ad))∫(dy/((ad−bc)y^2 +b))+(1/(ad))∫(dy/((1/c)−y^2 )) A=((ad−bc)/(ad(ad−bc)))∫(dy/(y^2 +[(√(b/(ad−bc)))]^2 ))+(1/(ad))∫(dy/(((1/(√c)))^2 −y^2 )) A=(1/(ad))(√((ad−bc)/b))tan^(−1) [((y(√(ad−bc)))/(√b))]+(1/(ad))×((√c)/2)ln[(1/((√c)/((1/(√c))−y)))+y ] A=(1/(ad))(√((ad−bc)/b))tan^(−1) [((y(√(ad−bc)))/(√b))]+((√c)/(2ad))ln[((1+y(√c))/(1−y(√c)))] A=(1/(ad))(√((ad−bc)/b))tan^(−1) [((t(√(ad−bc)))/(√(cbt^2 +bd)))]+((√c)/(2ad))ln[(((√(ct^2 +d))+t(√c))/((√(ct^2 +d))−t(√c)))]....(1) then B=∫((tdt)/((at^2 +b)(√(ct^2 +d))))=(1/2)∫((d(t^2 ))/((at^2 +b)(√(ct^2 +d)))) B=(1/2)∫(du/((au+b)(√(cu+d)))) ( put u=t^2 ,put p^2 =cu+d,2pdp=cdu) B=(1/(2c))∫((2pdp)/([a(((−d+p^2 )/c))+b]p))=(c/c)∫(dp/((ap^2 −ad+bc))) B=∫(dp/(ap^2 +(bc−ad)))=(1/a)∫(dp/(p^2 +(((bc−ad)/a)))) B=(1/a)∫(dp/(p^2 +((√((bc−ad)/a)))))=((√a)/(a(√(bc−ad))))tan^(−1) [((pdp)/(√(bc−ad)))] B=(1/((√a)(√(bc−ad))))tan^(−1) [((√(acu+da))/(√(bc−ad)))]+k B=(1/(√(abc−a^2 d)))tan^(−1) [((√(act^2 +ad ))/(√(bc−ad)))]+k.......(2) then I=−((2(√3))/(ad))(√((ad−bc)/b))tan^(−1) [((t(√(ad−bc)))/(√(cbt^2 +bd)))]−(((√3)(√c))/(ad))ln[(((√(ct^2 +d))+t(√c))/((√(ct^2 +d))−t(√c)))] −((2(√3))/(√(abc−a^2 d)))tan^(−1) [((√(act^2 +ad))/(√(bc−ad)))]+k but t=(((√3)−x)/((√3)+x)) I=−((2(√3))/(ad))(√((ad−bc)/b))tan^(−1) [((((√3)−x)(√(ad−bc)))/(((√3)+x)(√(cb((((√3)−x)/((√3)+x)))^2 +bd))))] −(((√3)(√c))/(ad))ln[(((√(c((((√3)−x)/((√3)+x)))^2 +d))+((((√3)−x)/((√3)+x)))(√c))/((√(c((((√3)−x)/((√3)+x)))^2 +d))−((((√3)−x)/((√3)+x)))(√c)))] −((2(√3))/(√(abc−a^2 d)))tan^(−1) [((√(ac((((√3)−x)/((√3)+x)))^2 +ad))/(√(bc−ad)))]+k where a=6−2(√3),b=6+2(√3),c=6−(√3),d=6+(√3) by mathdave (05/09/2020)

$${solution}\:{to}\: \\ $$$${let}\:{I}=\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}\:\:\:\:{put}\:{x}=\frac{\sqrt{\mathrm{3}}−\sqrt{\mathrm{3}}{t}}{\mathrm{1}+{t}}\:\:{and}\:\:{dx}=−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt} \\ $$$${I}=\int\frac{−\mathrm{2}\sqrt{\mathrm{3}}{dt}}{\left(\frac{\mathrm{3}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }+\mathrm{2}\sqrt{\mathrm{3}}\frac{\left(\mathrm{1}−{t}\right)}{\left(\mathrm{1}+{t}\right)}+\mathrm{3}\right)\sqrt{\frac{\mathrm{3}\left(\mathrm{1}−{t}\right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }+\sqrt{\mathrm{3}}\left(\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}\right)+\mathrm{3}}} \\ $$$${by}\:{simplify}\:{we}\:{have}\:{that} \\ $$$${I}=\int\frac{−\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+{t}\right){dt}}{\left.\left(\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} +\left(\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}}\right)\right)\sqrt{\left(\mathrm{6}−\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} +\left(\mathrm{6}+\sqrt{\mathrm{3}}\right)}} \\ $$$${let}\:{a}=\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}},{b}=\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}},{c}=\mathrm{6}−\sqrt{\mathrm{3}},{d}=\mathrm{6}+\sqrt{\mathrm{3}} \\ $$$$\int\frac{−\mathrm{2}\sqrt{\mathrm{3}}\left(\mathrm{1}+{t}\right){dt}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}=−\mathrm{2}\sqrt{\mathrm{3}}\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}−\mathrm{2}\sqrt{\mathrm{3}}\int\frac{{t}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}{dt} \\ $$$${Let}\:{A}=\int\frac{{dt}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}\:\:\:\:{let}\:\:{y}=\frac{{t}}{\sqrt{{ct}^{\mathrm{2}} +{d}}}\:\:,{t}^{\mathrm{2}} =\frac{{y}^{\mathrm{2}} {d}}{\mathrm{1}−{y}^{\mathrm{2}} {c}},{dt}=\frac{\sqrt{{d}}}{\left(\mathrm{1}−{y}^{\mathrm{2}} {c}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dy} \\ $$$${A}=\int\frac{{y}\sqrt{{d}}}{\frac{\left(\mathrm{1}−{y}^{\mathrm{2}} {c}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{\left[{a}\left(\frac{{y}^{\mathrm{2}} {d}}{\mathrm{1}−{y}^{\mathrm{2}} {c}}\right)+{b}\right]\sqrt{\frac{{y}^{\mathrm{2}} {d}}{\mathrm{1}−{y}^{\mathrm{2}} {c}}}}}{dy}= \\ $$$${A}=\int\frac{{dy}}{\left({ay}^{\mathrm{2}} {d}+{b}−{y}^{\mathrm{2}} {bc}\right)\left(\mathrm{1}−{y}^{\mathrm{2}} {c}\right)} \\ $$$${A}=\int\frac{\left({bc}−{ad}\right){dy}}{{ad}\left[\left({bc}−{ad}\right){y}^{\mathrm{2}} −{b}\right]}−\int\frac{{cdy}}{{ad}\left({cy}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$${A}=\frac{{ad}−{bc}}{{ad}}\int\frac{{dy}}{\left({ad}−{bc}\right){y}^{\mathrm{2}} +{b}}+\frac{\mathrm{1}}{{ad}}\int\frac{{dy}}{\frac{\mathrm{1}}{{c}}−{y}^{\mathrm{2}} } \\ $$$${A}=\frac{{ad}−{bc}}{{ad}\left({ad}−{bc}\right)}\int\frac{{dy}}{{y}^{\mathrm{2}} +\left[\sqrt{\frac{{b}}{{ad}−{bc}}}\right]^{\mathrm{2}} }+\frac{\mathrm{1}}{{ad}}\int\frac{{dy}}{\left(\frac{\mathrm{1}}{\sqrt{{c}}}\right)^{\mathrm{2}} −{y}^{\mathrm{2}} } \\ $$$${A}=\frac{\mathrm{1}}{{ad}}\sqrt{\frac{{ad}−{bc}}{{b}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{{y}\sqrt{{ad}−{bc}}}{\sqrt{{b}}}\right]+\frac{\mathrm{1}}{{ad}}×\frac{\sqrt{{c}}}{\mathrm{2}}\mathrm{ln}\left[\frac{\mathrm{1}}{\frac{\sqrt{{c}}}{\frac{\mathrm{1}}{\sqrt{{c}}}−{y}}}+{y}\:\right] \\ $$$${A}=\frac{\mathrm{1}}{{ad}}\sqrt{\frac{{ad}−{bc}}{{b}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{{y}\sqrt{{ad}−{bc}}}{\sqrt{{b}}}\right]+\frac{\sqrt{{c}}}{\mathrm{2}{ad}}\mathrm{ln}\left[\frac{\mathrm{1}+{y}\sqrt{{c}}}{\mathrm{1}−{y}\sqrt{{c}}}\right] \\ $$$${A}=\frac{\mathrm{1}}{{ad}}\sqrt{\frac{{ad}−{bc}}{{b}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{{t}\sqrt{{ad}−{bc}}}{\sqrt{{cbt}^{\mathrm{2}} +{bd}}}\right]+\frac{\sqrt{{c}}}{\mathrm{2}{ad}}\mathrm{ln}\left[\frac{\sqrt{{ct}^{\mathrm{2}} +{d}}+{t}\sqrt{{c}}}{\sqrt{{ct}^{\mathrm{2}} +{d}}−{t}\sqrt{{c}}}\right]....\left(\mathrm{1}\right) \\ $$$${then}\:{B}=\int\frac{{tdt}}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{d}\left({t}^{\mathrm{2}} \right)}{\left({at}^{\mathrm{2}} +{b}\right)\sqrt{{ct}^{\mathrm{2}} +{d}}} \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{du}}{\left({au}+{b}\right)\sqrt{{cu}+{d}}}\:\:\:\:\:\left(\:\:{put}\:{u}={t}^{\mathrm{2}} \:\:,{put}\:{p}^{\mathrm{2}} ={cu}+{d},\mathrm{2}{pdp}={cdu}\right) \\ $$$${B}=\frac{\mathrm{1}}{\mathrm{2}{c}}\int\frac{\mathrm{2}{pdp}}{\left[{a}\left(\frac{−{d}+{p}^{\mathrm{2}} }{{c}}\right)+{b}\right]{p}}=\frac{{c}}{{c}}\int\frac{{dp}}{\left({ap}^{\mathrm{2}} −{ad}+{bc}\right)} \\ $$$${B}=\int\frac{{dp}}{{ap}^{\mathrm{2}} +\left({bc}−{ad}\right)}=\frac{\mathrm{1}}{{a}}\int\frac{{dp}}{{p}^{\mathrm{2}} +\left(\frac{{bc}−{ad}}{{a}}\right)} \\ $$$${B}=\frac{\mathrm{1}}{{a}}\int\frac{{dp}}{{p}^{\mathrm{2}} +\left(\sqrt{\frac{{bc}−{ad}}{{a}}}\right)}=\frac{\sqrt{{a}}}{{a}\sqrt{{bc}−{ad}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{{pdp}}{\sqrt{{bc}−{ad}}}\right] \\ $$$${B}=\frac{\mathrm{1}}{\sqrt{{a}}\sqrt{{bc}−{ad}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{{acu}+{da}}}{\sqrt{{bc}−{ad}}}\right]+{k} \\ $$$${B}=\frac{\mathrm{1}}{\sqrt{{abc}−{a}^{\mathrm{2}} {d}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{{act}^{\mathrm{2}} +{ad}\:}}{\sqrt{{bc}−{ad}}}\right]+{k}.......\left(\mathrm{2}\right) \\ $$$${then}\: \\ $$$${I}=−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{ad}}\sqrt{\frac{{ad}−{bc}}{{b}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{{t}\sqrt{{ad}−{bc}}}{\sqrt{{cbt}^{\mathrm{2}} +{bd}}}\right]−\frac{\sqrt{\mathrm{3}}\sqrt{{c}}}{{ad}}\mathrm{ln}\left[\frac{\sqrt{{ct}^{\mathrm{2}} +{d}}+{t}\sqrt{{c}}}{\sqrt{{ct}^{\mathrm{2}} +{d}}−{t}\sqrt{{c}}}\right] \\ $$$$\:\:\:\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\sqrt{{abc}−{a}^{\mathrm{2}} {d}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{{act}^{\mathrm{2}} +{ad}}}{\sqrt{{bc}−{ad}}}\right]+{k} \\ $$$${but}\:{t}=\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}} \\ $$$${I}=−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{{ad}}\sqrt{\frac{{ad}−{bc}}{{b}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\left(\sqrt{\mathrm{3}}−{x}\right)\sqrt{{ad}−{bc}}}{\left(\sqrt{\mathrm{3}}+{x}\right)\sqrt{{cb}\left(\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}}\right)^{\mathrm{2}} +{bd}}}\right] \\ $$$$\:\:\:\:\:−\frac{\sqrt{\mathrm{3}}\sqrt{{c}}}{{ad}}\mathrm{ln}\left[\frac{\sqrt{{c}\left(\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}}\right)^{\mathrm{2}} +{d}}+\left(\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}}\right)\sqrt{{c}}}{\sqrt{{c}\left(\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}}\right)^{\mathrm{2}} +{d}}−\left(\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}}\right)\sqrt{{c}}}\right] \\ $$$$\:\:\:\:\:\:−\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\sqrt{{abc}−{a}^{\mathrm{2}} {d}}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{\sqrt{{ac}\left(\frac{\sqrt{\mathrm{3}}−{x}}{\sqrt{\mathrm{3}}+{x}}\right)^{\mathrm{2}} +{ad}}}{\sqrt{{bc}−{ad}}}\right]+{k} \\ $$$${where}\:{a}=\mathrm{6}−\mathrm{2}\sqrt{\mathrm{3}},{b}=\mathrm{6}+\mathrm{2}\sqrt{\mathrm{3}},{c}=\mathrm{6}−\sqrt{\mathrm{3}},{d}=\mathrm{6}+\sqrt{\mathrm{3}} \\ $$$${by}\:{mathdave}\:\left(\mathrm{05}/\mathrm{09}/\mathrm{2020}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Tawa11 last updated on 06/Sep/21

great sir

$$\mathrm{great}\:\mathrm{sir} \\ $$

Answered by ajfour last updated on 06/Sep/20

I=∫(dx/({(x+1)^2 +2}(√((x+(1/2))^2 +((11)/4))))) let x+(1/2)=((√(11))/2)tan θ I=∫((((√(11))/2)sec^2 θdθ)/({((((√(11))tan θ+1)^2 )/4)+2}((√(11))/2)sec θ)) =4∫((sec θdθ)/(((√(11))tan θ+1)^2 +8)) =4∫((cos θdθ)/(11sin^2 θ+(√(11))sin 2θ+9cos^2 θ)) =4∫((cos θdθ)/(11+(√(11))sin 2θ−1−cos 2θ)) I =4∫((cos θdθ)/(10−2(√3)cos (2θ+tan^(−1) (√(11))))) say θ+β=φ ; β=(1/2)tan^(−1) (√(11)) I=4∫((cos (φ−β)dφ)/(10−2(√3)cos (2φ))) I =4cos β∫((cos φdφ)/(10−2(√3)+4(√3)sin^2 φ)) −4sin β∫(((−sin φ)dφ)/(10+2(√3)−4(√3)cos^2 φ))+c −−−−−−−−−−−−−−−−−− I=((cos β)/( (√3)))×(√((4(√3))/(10−2(√3))))tan^(−1) [(√((4(√3))/(10−2(√3))))sin (θ+β)] −((sin β)/( (√3)))×(1/2)(√((4(√3))/(10+2(√3))))ln ∣(((√((10+2(√3))/(4(√3))))+cos (θ+β))/( (√((10+2(√3))/(4(√3))))−cos (θ+β)))∣+c θ=tan^(−1) (((2x+1)/( (√(11))))) ; β=(1/2)tan^(−1) (√(11)) ★−−−−−−−−−−−−−−−−★ rough work tan 2β=((2tan β)/(1−tan^2 β))=(√(11)) ⇒ (√(11))tan^2 β+2tan β−(√(11))=0 tan β=((2(√3)−1)/( (√(11)))) sin β=((2(√3)−1)/( (√(24−4(√3))))) = (√((2(√3)−1)/(4(√3)))) cos β= ((√(11))/( (√(4(√3)))(√(2(√3)−1)))) _____________________________ I=(((3(√2)+(√6))/(12)))tan^(−1) [(√((4(√3))/(10−2(√3))))sin φ] −((√(3(√3)−3))/(12)) ln ∣(((√(10+2(√3)))+(√(4(√3)))cos φ)/( (√(10+2(√3)))−(√(4(√3)))cos φ))∣+c ............................................. ∀ tan φ=(((2(√3)+1)/( (√(11)))))(((x+(√3))/( (√3)−x))) _____________________________ rough work tan (θ+β)=((((2x+1)/( (√(11))))+((2(√3)−1)/( (√(11)))))/(1−(((2x+1)(2(√3)−1))/(11)))) =(((√(11))(x+(√3)))/(6−(√3)−(2(√3)−1)x)) =(((√(11))(x+(√3)))/((2(√3)−1)((√3)−x))) −−−−−−−−−−−−−−

$${I}=\int\frac{{dx}}{\left\{\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}\right\}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{11}}{\mathrm{4}}}} \\ $$$${let}\:\:\:{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{tan}\:\theta \\ $$$${I}=\int\frac{\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \theta{d}\theta}{\left\{\frac{\left(\sqrt{\mathrm{11}}\mathrm{tan}\:\theta+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\mathrm{2}\right\}\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{sec}\:\theta} \\ $$$$\:\:\:=\mathrm{4}\int\frac{\mathrm{sec}\:\theta{d}\theta}{\left(\sqrt{\mathrm{11}}\mathrm{tan}\:\theta+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{8}} \\ $$$$\:\:\:=\mathrm{4}\int\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{11sin}\:^{\mathrm{2}} \theta+\sqrt{\mathrm{11}}\mathrm{sin}\:\mathrm{2}\theta+\mathrm{9cos}\:^{\mathrm{2}} \theta} \\ $$$$\:\:=\mathrm{4}\int\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{11}+\sqrt{\mathrm{11}}\mathrm{sin}\:\mathrm{2}\theta−\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta} \\ $$$$\:{I}\:=\mathrm{4}\int\frac{\mathrm{cos}\:\theta{d}\theta}{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{cos}\:\left(\mathrm{2}\theta+\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{11}}\right)} \\ $$$$\:\:{say}\:\:\:\theta+\beta=\phi\:\:;\:\:\:\beta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{11}} \\ $$$${I}=\mathrm{4}\int\frac{\mathrm{cos}\:\left(\phi−\beta\right){d}\phi}{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}\mathrm{cos}\:\left(\mathrm{2}\phi\right)} \\ $$$$\:{I}\:=\mathrm{4cos}\:\beta\int\frac{\mathrm{cos}\:\phi{d}\phi}{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{4}\sqrt{\mathrm{3}}\mathrm{sin}\:^{\mathrm{2}} \phi} \\ $$$$\:\:\:\:\:\:\:\:−\mathrm{4sin}\:\beta\int\frac{\left(−\mathrm{sin}\:\phi\right){d}\phi}{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{4}\sqrt{\mathrm{3}}\mathrm{cos}\:^{\mathrm{2}} \phi}+{c} \\ $$$$−−−−−−−−−−−−−−−−−− \\ $$$${I}=\frac{\mathrm{cos}\:\beta}{\:\sqrt{\mathrm{3}}}×\sqrt{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}}}\mathrm{tan}^{−\mathrm{1}} \left[\sqrt{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}}}\mathrm{sin}\:\left(\theta+\beta\right)\right] \\ $$$$\:\:\:−\frac{\mathrm{sin}\:\beta}{\:\sqrt{\mathrm{3}}}×\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}}}\mathrm{ln}\:\mid\frac{\sqrt{\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}}}+\mathrm{cos}\:\left(\theta+\beta\right)}{\:\sqrt{\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{4}\sqrt{\mathrm{3}}}}−\mathrm{cos}\:\left(\theta+\beta\right)}\mid+{c} \\ $$$$\:\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)\:\:;\:\:\beta=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{11}} \\ $$$$\bigstar−−−−−−−−−−−−−−−−\bigstar \\ $$$${rough}\:{work} \\ $$$$\mathrm{tan}\:\mathrm{2}\beta=\frac{\mathrm{2tan}\:\beta}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} \beta}=\sqrt{\mathrm{11}} \\ $$$$\Rightarrow\:\:\:\sqrt{\mathrm{11}}\mathrm{tan}\:^{\mathrm{2}} \beta+\mathrm{2tan}\:\beta−\sqrt{\mathrm{11}}=\mathrm{0} \\ $$$$\mathrm{tan}\:\beta=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{11}}} \\ $$$$\mathrm{sin}\:\beta=\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{24}−\mathrm{4}\sqrt{\mathrm{3}}}}\:=\:\sqrt{\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}} \\ $$$$\mathrm{cos}\:\beta=\:\frac{\sqrt{\mathrm{11}}}{\:\sqrt{\mathrm{4}\sqrt{\mathrm{3}}}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${I}=\left(\frac{\mathrm{3}\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{12}}\right)\mathrm{tan}^{−\mathrm{1}} \left[\sqrt{\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{3}}}}\mathrm{sin}\:\phi\right] \\ $$$$\:−\frac{\sqrt{\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}}}{\mathrm{12}}\:\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}}+\sqrt{\mathrm{4}\sqrt{\mathrm{3}}}\mathrm{cos}\:\phi}{\:\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{3}}}−\sqrt{\mathrm{4}\sqrt{\mathrm{3}}}\mathrm{cos}\:\phi}\mid+{c} \\ $$$$\:\:\:\:\:\:\:............................................. \\ $$$$\:\:\:\forall\:\:\:\:\mathrm{tan}\:\phi=\left(\frac{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)\left(\frac{{x}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{3}}−{x}}\right) \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${rough}\:{work} \\ $$$$\mathrm{tan}\:\left(\theta+\beta\right)=\frac{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}+\frac{\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}}{\:\sqrt{\mathrm{11}}}}{\mathrm{1}−\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)}{\mathrm{11}}} \\ $$$$\:\:=\frac{\sqrt{\mathrm{11}}\left({x}+\sqrt{\mathrm{3}}\right)}{\mathrm{6}−\sqrt{\mathrm{3}}−\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right){x}}\:=\frac{\sqrt{\mathrm{11}}\left({x}+\sqrt{\mathrm{3}}\right)}{\left(\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{1}\right)\left(\sqrt{\mathrm{3}}−{x}\right)} \\ $$$$\:\:\:\:\:\:\:−−−−−−−−−−−−−− \\ $$

Commented by ajfour last updated on 06/Sep/20

Isn′t this alright ?

$${Isn}'{t}\:\:{this}\:{alright}\:? \\ $$

Commented by MJS_new last updated on 06/Sep/20

I′ll check it tomorrow

$$\mathrm{I}'\mathrm{ll}\:\mathrm{check}\:\mathrm{it}\:\mathrm{tomorrow} \\ $$

Commented by mathdave last updated on 06/Sep/20

this cant b true

$${this}\:{cant}\:{b}\:{true} \\ $$

Commented by MJS_new last updated on 06/Sep/20

it′s right, Sir Aifour. can you finish it?

$$\mathrm{it}'\mathrm{s}\:\mathrm{right},\:\mathrm{Sir}\:\mathrm{Aifour}.\:\mathrm{can}\:\mathrm{you}\:\mathrm{finish}\:\mathrm{it}? \\ $$

Commented by ajfour last updated on 06/Sep/20

thank you Sir!

$${thank}\:{you}\:{Sir}! \\ $$

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