Question Number 36545 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$$\int\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx} \\ $$
Commented by MJS last updated on 03/Jun/18
$$…\mathrm{I}\:\mathrm{just}\:\mathrm{gave}\:\mathrm{it}\:\mathrm{a}\:\mathrm{try}\:\mathrm{because}\:\mathrm{the}\:“\mathrm{poper}''\: \\ $$$$\mathrm{way}\:\mathrm{looked}\:\mathrm{rather}\:\mathrm{complicated}… \\ $$
Commented by MJS last updated on 03/Jun/18
$$\mathrm{this}\:\mathrm{looks}\:\mathrm{like}\:\mathrm{it}'\mathrm{s}\:\mathrm{going}\:\mathrm{to}\:\mathrm{be} \\ $$$$\frac{{ax}+{b}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}} \\ $$$$ \\ $$$$\left(\frac{{ax}+{b}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}}\right)'= \\ $$$$=\frac{\left(\mathrm{2}{a}−{b}\right){x}+\left({a}−\mathrm{2}{b}\right)}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\overset{\frac{\mathrm{3}}{\mathrm{2}}} {\right)}} \\ $$$$\mathrm{2}{a}−{b}=\mathrm{2} \\ $$$${a}−\mathrm{2}{b}=\mathrm{1} \\ $$$${a}=\mathrm{1} \\ $$$${b}=\mathrm{0} \\ $$$$ \\ $$$$\int\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}}+{C} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$${excellent}\:{sir}… \\ $$
Commented by ajfour last updated on 03/Jun/18
$${couldn}'{t}\:{follow}…?? \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18
$${sir}\:{mjs}\:{assumed}\:{the}\:{results}\:{of}\:{the}\:{intregal} \\ $$$${as}\:\frac{{ax}+{b}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}\:\:}{then}\:{differentiated}\:{and}\:{equatex} \\ $$$${and}\:{gotthe}\:{results} \\ $$
Commented by ajfour last updated on 03/Jun/18
$${thanks},\:{i}\:{see}. \\ $$
Answered by ajfour last updated on 03/Jun/18
![=∫((2x+1)/([(x+2)^2 −((√3))^2 ]^(3/2) ))dx let x+2=(√3)sec θ ⇒ dx=(√3)sec θtan θdθ I=∫((2(√3)sec θ−3)/(3(√3)sec^3 θ))dθ =(2/3)∫cos^2 θdθ−(1/( (√3)))∫cos^3 θdθ =(1/3)(θ+((sin 2θ)/2))−(1/(4(√3)))∫(cos 3θ+3cos θ)dθ =(θ/3)+((sin 2θ)/6)−((sin 3θ)/(12(√3)))−((√3)/4)sin θ+c .](https://www.tinkutara.com/question/Q36562.png)
$$=\int\frac{\mathrm{2}{x}+\mathrm{1}}{\left[\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \right]^{\mathrm{3}/\mathrm{2}} }{dx} \\ $$$${let}\:\:{x}+\mathrm{2}=\sqrt{\mathrm{3}}\mathrm{sec}\:\theta \\ $$$$\Rightarrow\:\:\:\:\:{dx}=\sqrt{\mathrm{3}}\mathrm{sec}\:\theta\mathrm{tan}\:\theta{d}\theta \\ $$$${I}=\int\frac{\mathrm{2}\sqrt{\mathrm{3}}\mathrm{sec}\:\theta−\mathrm{3}}{\mathrm{3}\sqrt{\mathrm{3}}\mathrm{sec}\:^{\mathrm{3}} \theta}{d}\theta \\ $$$$\:\:\:=\frac{\mathrm{2}}{\mathrm{3}}\int\mathrm{cos}\:^{\mathrm{2}} \theta{d}\theta−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\mathrm{cos}\:^{\mathrm{3}} \theta{d}\theta \\ $$$$\:\:=\frac{\mathrm{1}}{\mathrm{3}}\left(\theta+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{3}}}\int\left(\mathrm{cos}\:\mathrm{3}\theta+\mathrm{3cos}\:\theta\right){d}\theta \\ $$$$\:=\frac{\theta}{\mathrm{3}}+\frac{\mathrm{sin}\:\mathrm{2}\theta}{\mathrm{6}}−\frac{\mathrm{sin}\:\mathrm{3}\theta}{\mathrm{12}\sqrt{\mathrm{3}}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\mathrm{sin}\:\theta+{c}\:. \\ $$
Answered by MJS last updated on 03/Jun/18
![∫((2x+1)/((x^2 +4x+1)^(3/2) ))dx=∫((2x+4)/((x^2 +4x+1)^(3/2) ))dx−3∫(dx/((x^2 +4x+1)^(3/2) ))= [((∫((2x+4)/((x^2 +4x+1)^(3/2) ))dx=)),(( [t=x^2 +4x+1 → dx=(dt/(2x+4))])),((=∫(dt/t^(3/2) )=−(2/( (√t)))=−(2/( (√(x^2 +4x+1)))))) ] [((∫(dx/((x^2 +4x+1)^(3/2) ))=∫(dx/(((x+2)^2 −3)^(3/2) ))=)),(( [u=x+2 → dx=du])),((=∫(du/((u^2 −3)^(3/2) ))=)),(( [v=(√3)sec v → du=(√3)sec v tan v dv])),((=∫(((√3)sec v tan v)/((3sec^2 v−3)^(3/2) ))dv=(1/3)∫((sec v)/(tan^2 v))dv=(1/3)∫((cos v)/(sin^2 v))dv=)),(( [w=sin v → dv=(dw/(cos v))])),((=(1/3)∫(dw/w^2 )=−(1/(3w))=−(1/(3sin v))=−(u/(3(√(u^2 −3))))=)),((=−((x+2)/(3(√(x^2 +4x+1)))))) ] =(x/( (√(x^2 +4x+1))))+C](https://www.tinkutara.com/question/Q36636.png)
$$\int\frac{\mathrm{2}{x}+\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=\int\frac{\mathrm{2}{x}+\mathrm{4}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}−\mathrm{3}\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\int\frac{\mathrm{2}{x}+\mathrm{4}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dx}=}\\{\:\:\:\:\:\:\:\:\:\:\left[{t}={x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\:\rightarrow\:{dx}=\frac{{dt}}{\mathrm{2}{x}+\mathrm{4}}\right]}\\{=\int\frac{{dt}}{{t}^{\frac{\mathrm{3}}{\mathrm{2}}} }=−\frac{\mathrm{2}}{\:\sqrt{{t}}}=−\frac{\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}}}\end{bmatrix} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\begin{bmatrix}{\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=\int\frac{{dx}}{\left(\left({x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=}\\{\:\:\:\:\:\:\:\:\:\:\left[{u}={x}+\mathrm{2}\:\rightarrow\:{dx}={du}\right]}\\{=\int\frac{{du}}{\left({u}^{\mathrm{2}} −\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }=}\\{\:\:\:\:\:\:\:\:\:\:\left[{v}=\sqrt{\mathrm{3}}\mathrm{sec}\:{v}\:\rightarrow\:{du}=\sqrt{\mathrm{3}}\mathrm{sec}\:{v}\:\mathrm{tan}\:{v}\:{dv}\right]}\\{=\int\frac{\sqrt{\mathrm{3}}\mathrm{sec}\:{v}\:\mathrm{tan}\:{v}}{\left(\mathrm{3sec}^{\mathrm{2}} \:{v}−\mathrm{3}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }{dv}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{sec}\:{v}}{\mathrm{tan}^{\mathrm{2}} \:{v}}{dv}=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{cos}\:{v}}{\mathrm{sin}^{\mathrm{2}} \:{v}}{dv}=}\\{\:\:\:\:\:\:\:\:\:\:\left[{w}=\mathrm{sin}\:{v}\:\rightarrow\:{dv}=\frac{{dw}}{\mathrm{cos}\:{v}}\right]}\\{=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dw}}{{w}^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{3}{w}}=−\frac{\mathrm{1}}{\mathrm{3sin}\:{v}}=−\frac{{u}}{\mathrm{3}\sqrt{{u}^{\mathrm{2}} −\mathrm{3}}}=}\\{=−\frac{{x}+\mathrm{2}}{\mathrm{3}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}}}\end{bmatrix} \\ $$$$ \\ $$$$=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{1}}}+{C} \\ $$