Question-49251

Question Number 49251 by cesar.marval.larez@gmail.com last updated on 04/Dec/18

Commented by afachri last updated on 05/Dec/18

unfortunately i am not learning integral yet

$$\mathrm{unfortunately}\:\mathrm{i}\:\mathrm{am}\:\mathrm{not}\:\mathrm{learning}\:\mathrm{integral} \\ $$$$\mathrm{yet} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

1)∫_0 ^1 ((sin(((5x)/3)))/(cos^2 (((5x)/3))))dx t=cos(((5x)/3)) so dt=−sin(((5x)/3))×(5/3)dx ((−3)/5)dt=sin(((5x)/3))dx =((−3)/5)∫_1 ^(cos((5/3))) (dt/t^2 ) =((−3)/5)×∣((−1)/t)∣_1 ^(cos((5/3))) =(3/5){(1/(cos((5/3))))−1}

$$\left.\mathrm{1}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{sin}\left(\frac{\mathrm{5}{x}}{\mathrm{3}}\right)}{{cos}^{\mathrm{2}} \left(\frac{\mathrm{5}{x}}{\mathrm{3}}\right)}{dx} \\ $$$${t}={cos}\left(\frac{\mathrm{5}{x}}{\mathrm{3}}\right)\:\:{so}\:{dt}=−{sin}\left(\frac{\mathrm{5}{x}}{\mathrm{3}}\right)×\frac{\mathrm{5}}{\mathrm{3}}{dx} \\ $$$$\frac{−\mathrm{3}}{\mathrm{5}}{dt}={sin}\left(\frac{\mathrm{5}{x}}{\mathrm{3}}\right){dx} \\ $$$$=\frac{−\mathrm{3}}{\mathrm{5}}\int_{\mathrm{1}} ^{{cos}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)} \:\frac{{dt}}{{t}^{\mathrm{2}} } \\ $$$$=\frac{−\mathrm{3}}{\mathrm{5}}×\mid\frac{−\mathrm{1}}{{t}}\mid_{\mathrm{1}} ^{{cos}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)} \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}\left\{\frac{\mathrm{1}}{{cos}\left(\frac{\mathrm{5}}{\mathrm{3}}\right)}−\mathrm{1}\right\} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

2)∫_0 ^1 e^((3x)/5) cos(((4x)/3)) dx let we find step 1−find ∫e^(ax) cosbx dx then put a=(3/5) and b=(4/3) finally put upper and lower limit method−1 I=∫e^(ax) cosbxdx =e^(ax) ×((sinbx)/b)−∫[(d/dx)(e^(ax) )∫cosbxdx]dx =((e^(ax) sinbx)/b)−∫ae^(ax) ×((sinbx)/b)dx =((e^(ax) sinbx)/b)−(a/b)[e^(ax) ∫sinbxdx−∫{((d(e^(ax) ))/dx)∫sinbxdx}dx] =do−(a/b)[e^(ax) ×((−cosbx)/b)−∫ae^(ax) ×((−cosbx)/b)dx] =do−(a/b)[((−e^(ax) cosbx)/b)+(a/b)∫e^(ax) cosbxdx] =do−(a/b)[((−e^(ax) cosbx)/b)+(a/b)I] =((e^(ax) sinbx)/b)+(a/b^2 )e^(ax) cosbx−(a^2 /b^2 )I I+(a^2 /b^2 )I=((be^(ax) sinbx+ae^(ax) cosbx)/b^2 ) I(1+(a^2 /b^2 ))=((e^(ax) (bsinbx+acosbx))/b^2 ) I(((a^2 +b^2 )/b^2 ))=((e^(ax) (bsinbx+acosbx))/b^2 ) I=((e^(ax) (bsinbx+acosbx))/(a^2 +b^2 )) now put a=(3/5) b=(4/3) so required intregal is ∣((e^((3/5)x) {((4/3)sin((4/3)x)+(3/5)cos((4/3)x)})/(((3/5))^2 +((4/3))^2 ))∣_0 ^1 =[((e^(3/5) {(4/3)sin((4/3))+(3/5)cos((4/3))})/(((3/5))^2 +((4/3))^2 ))]−[((e^0 {(4/3)sin0+(3/5)cos(0))/(((3/5))^2 +((4/3))^2 ))] =[(e^((3/5){(4/3)sin((4/3))+(3/5)cos((4/3))}−(3/5)) /(((3/5))^2 +((4/3))^2 ))]pls check

$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\frac{\mathrm{3}{x}}{\mathrm{5}}} {cos}\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)\:{dx} \\ $$$${let}\:{we}\:{find}\:\:{step}\:\mathrm{1}−{find}\:\int{e}^{{ax}} {cosbx}\:{dx} \\ $$$$ \\ $$$${then}\:{put}\:{a}=\frac{\mathrm{3}}{\mathrm{5}}\:{and}\:{b}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${finally}\:{put}\:{upper}\:{and}\:{lower}\:{limit} \\ $$$${method}−\mathrm{1} \\ $$$${I}=\int{e}^{{ax}} {cosbxdx} \\ $$$$={e}^{{ax}} ×\frac{{sinbx}}{{b}}−\int\left[\frac{{d}}{{dx}}\left({e}^{{ax}} \right)\int{cosbxdx}\right]{dx} \\ $$$$=\frac{{e}^{{ax}} {sinbx}}{{b}}−\int{ae}^{{ax}} ×\frac{{sinbx}}{{b}}{dx} \\ $$$$=\frac{{e}^{{ax}} {sinbx}}{{b}}−\frac{{a}}{{b}}\left[{e}^{{ax}} \int{sinbxdx}−\int\left\{\frac{{d}\left({e}^{{ax}} \right)}{{dx}}\int{sinbxdx}\right\}{dx}\right] \\ $$$$={do}−\frac{{a}}{{b}}\left[{e}^{{ax}} ×\frac{−{cosbx}}{{b}}−\int{ae}^{{ax}} ×\frac{−{cosbx}}{{b}}{dx}\right] \\ $$$$={do}−\frac{{a}}{{b}}\left[\frac{−{e}^{{ax}} {cosbx}}{{b}}+\frac{{a}}{{b}}\int{e}^{{ax}} {cosbxdx}\right] \\ $$$$={do}−\frac{{a}}{{b}}\left[\frac{−{e}^{{ax}} {cosbx}}{{b}}+\frac{{a}}{{b}}{I}\right] \\ $$$$=\frac{{e}^{{ax}} {sinbx}}{{b}}+\frac{{a}}{{b}^{\mathrm{2}} }{e}^{{ax}} {cosbx}−\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{I} \\ $$$${I}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }{I}=\frac{{be}^{{ax}} {sinbx}+{ae}^{{ax}} {cosbx}}{{b}^{\mathrm{2}} } \\ $$$${I}\left(\mathrm{1}+\frac{{a}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)=\frac{{e}^{{ax}} \left({bsinbx}+{acosbx}\right)}{{b}^{\mathrm{2}} } \\ $$$${I}\left(\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right)=\frac{{e}^{{ax}} \left({bsinbx}+{acosbx}\right)}{{b}^{\mathrm{2}} } \\ $$$${I}=\frac{{e}^{{ax}} \left({bsinbx}+{acosbx}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${now}\:{put}\:{a}=\frac{\mathrm{3}}{\mathrm{5}}\:\:\:{b}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${so}\:{required}\:{intregal}\:{is} \\ $$$$\mid\frac{{e}^{\frac{\mathrm{3}}{\mathrm{5}}{x}} \left\{\left(\frac{\mathrm{4}}{\mathrm{3}}{sin}\left(\frac{\mathrm{4}}{\mathrm{3}}{x}\right)+\frac{\mathrm{3}}{\mathrm{5}}{cos}\left(\frac{\mathrm{4}}{\mathrm{3}}{x}\right)\right\}\right.}{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} }\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\left[\frac{{e}^{\frac{\mathrm{3}}{\mathrm{5}}} \left\{\frac{\mathrm{4}}{\mathrm{3}}{sin}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+\frac{\mathrm{3}}{\mathrm{5}}{cos}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right\}}{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} }\right]−\left[\frac{{e}^{\mathrm{0}} \left\{\frac{\mathrm{4}}{\mathrm{3}}{sin}\mathrm{0}+\frac{\mathrm{3}}{\mathrm{5}}{cos}\left(\mathrm{0}\right)\right.}{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} }\right] \\ $$$$=\left[\frac{{e}^{\frac{\mathrm{3}}{\mathrm{5}}\left\{\frac{\mathrm{4}}{\mathrm{3}}{sin}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)+\frac{\mathrm{3}}{\mathrm{5}}{cos}\left(\frac{\mathrm{4}}{\mathrm{3}}\right)\right\}−\frac{\mathrm{3}}{\mathrm{5}}} }{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} }\right]{pls}\:{check}\: \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

Commented by cesar.marval.larez@gmail.com last updated on 05/Dec/18

$${Woooow}\:{my}\:{friend}\:{i}\:{am}\:{without}\:{word} \\ $$$${thank}\:{u}\:{very}\:{much}\:{and}\:{congragulations} \\ $$$${for}\:{this}\:{talent}.\:{i}\:{am}\:{very}\:{grateful}.\: \\ $$$${where}\:{are}\:{u}\:{from}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

$${india}\:\:{nagpur} \\ $$

Commented by cesar.marval.larez@gmail.com last updated on 05/Dec/18

$${U}\:{know}\:{much}\:{u}\:{teach}\:{with}\:{this}\:{work}. \\ $$$${have}\:{u}\:{some}\:{social}\:{conextion}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

$$\:{i}\:{have}\:{facebook}\:{accout}\: \\ $$

Commented by cesar.marval.larez@gmail.com last updated on 06/Dec/18

$${may}\:{i}\:{add}\:{u}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Dec/18

$${ok}\:{you}\:{can}\:{add}\:{me} \\ $$

Commented by cesar.marval.larez@gmail.com last updated on 07/Dec/18

$${Whats}\:{your}\:{name}? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18

$${TANMAY}\:\:{CHAUDHURY} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

2)∫_0 ^1 e^((3x)/5) cos((4/3))dx altetrnative method... ∫e^(ax) cosbx dx a=(3/5) b=(4/3) p=∫e^(ax) cosbxdx q=∫e^(ax) sinbxdx p+iq=∫e^(ax) (cosbx+isinbx)dx =∫e^(ax) ×e^(ibx) dx=∫e^((a+ib)x) dx p+iq=(e^((a+ib)x) /(a+ib)) p+iq=((e^(ax) ×e^(ibx) )/((a+ib)(a−ib)))×(a−ib) =(e^(ax) /(a^2 +b^2 ))(cosbx+isinbx)(a−ib) formula e^(iα) =cosα+isinα =(e^(ax) /(a^2 +b^2 ))×(acosbx+iasinbx−ibcosbx+bsinbx) =(e^(ax) /(a^2 +b^2 ))(acosbx+bsinbx)+i×(e^(ax) /(a^2 +b^2 ))(asinbx−bcosbx) so equating real and imaginary part both side p=∫e^(ax) cosbxdx=(e^(ax) /(a^2 +b^2 ))(acosbx+bsinbx) q=∫e^(ax) sinbxdx=(e^(ax) /(a^2 +b^2 ))(asinbx−bcosbx) now ∫_0 ^1 e^((3x)/5) cos(((4x)/3))dx ∣(e^((3x)/5) /(((3/5))^2 +((4/3))^2 ))×{(3/5)cos(((4x)/3))+(4/3)×sin(((4x)/3))}∣_0 ^1 already solved..

$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\frac{\mathrm{3}{x}}{\mathrm{5}}} {cos}\left(\frac{\mathrm{4}}{\mathrm{3}}\right){dx} \\ $$$${altetrnative}\:{method}… \\ $$$$\int{e}^{{ax}} {cosbx}\:{dx}\:\:\:{a}=\frac{\mathrm{3}}{\mathrm{5}}\:\:{b}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$$${p}=\int{e}^{{ax}} {cosbxdx} \\ $$$${q}=\int{e}^{{ax}} {sinbxdx} \\ $$$${p}+{iq}=\int{e}^{{ax}} \left({cosbx}+{isinbx}\right){dx} \\ $$$$=\int{e}^{{ax}} ×{e}^{{ibx}} {dx}=\int{e}^{\left({a}+{ib}\right){x}} {dx} \\ $$$${p}+{iq}=\frac{{e}^{\left({a}+{ib}\right){x}} }{{a}+{ib}} \\ $$$${p}+{iq}=\frac{{e}^{{ax}} ×{e}^{{ibx}} }{\left({a}+{ib}\right)\left({a}−{ib}\right)}×\left({a}−{ib}\right) \\ $$$$=\frac{{e}^{{ax}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({cosbx}+{isinbx}\right)\left({a}−{ib}\right) \\ $$$${formula}\:{e}^{{i}\alpha} ={cos}\alpha+{isin}\alpha \\ $$$$=\frac{{e}^{{ax}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }×\left({acosbx}+{iasinbx}−{ibcosbx}+{bsinbx}\right) \\ $$$$=\frac{{e}^{{ax}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({acosbx}+{bsinbx}\right)+{i}×\frac{{e}^{{ax}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({asinbx}−{bcosbx}\right) \\ $$$${so}\:{equating}\:{real}\:{and}\:{imaginary}\:{part}\:{both}\:{side} \\ $$$${p}=\int{e}^{{ax}} {cosbxdx}=\frac{{e}^{{ax}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({acosbx}+{bsinbx}\right) \\ $$$${q}=\int{e}^{{ax}} {sinbxdx}=\frac{{e}^{{ax}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\left({asinbx}−{bcosbx}\right) \\ $$$${now}\:\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{\frac{\mathrm{3}{x}}{\mathrm{5}}} {cos}\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right){dx} \\ $$$$\mid\frac{{e}^{\frac{\mathrm{3}{x}}{\mathrm{5}}} }{\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{4}}{\mathrm{3}}\right)^{\mathrm{2}} }×\left\{\frac{\mathrm{3}}{\mathrm{5}}{cos}\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)+\frac{\mathrm{4}}{\mathrm{3}}×{sin}\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)\right\}\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$${already}\:{solved}.. \\ $$$$ \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

3)∫sin^5 (3x)cos^2 (3x)dx ∫sin(3x){1−cos^2 (3x)}^2 cos^2 (3x)dx t=cos(3x) dt=−3sin(3x)dx (dt/(−3))=sin(3x)dx ∫(1−t^2 )^2 ×t^2 ×(dt/(−3)) ((−1)/3)∫(1−2t^2 +t^4 )t^2 dt =((−1)/3)∫(t^2 −2t^4 +t^6 )dt =((−1)/3)[(t^3 /3)−((2t^5 )/5)+(t^7 /7)]+c =((−1)/3)[((cos^3 (3x))/3)−((2cos^5 (3x))/5)+((cos^7 (3x))/7)]+c

$$\left.\mathrm{3}\right)\int{sin}^{\mathrm{5}} \left(\mathrm{3}{x}\right){cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){dx} \\ $$$$\int{sin}\left(\mathrm{3}{x}\right)\left\{\mathrm{1}−{cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right)\right\}^{\mathrm{2}} {cos}^{\mathrm{2}} \left(\mathrm{3}{x}\right){dx} \\ $$$${t}={cos}\left(\mathrm{3}{x}\right) \\ $$$${dt}=−\mathrm{3}{sin}\left(\mathrm{3}{x}\right){dx} \\ $$$$\frac{{dt}}{−\mathrm{3}}={sin}\left(\mathrm{3}{x}\right){dx} \\ $$$$\int\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} ×{t}^{\mathrm{2}} ×\frac{{dt}}{−\mathrm{3}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{3}}\int\left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{2}} +{t}^{\mathrm{4}} \right){t}^{\mathrm{2}} {dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}\int\left({t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{4}} +{t}^{\mathrm{6}} \right){dt} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}\left[\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−\frac{\mathrm{2}{t}^{\mathrm{5}} }{\mathrm{5}}+\frac{{t}^{\mathrm{7}} }{\mathrm{7}}\right]+{c} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}\left[\frac{{cos}^{\mathrm{3}} \left(\mathrm{3}{x}\right)}{\mathrm{3}}−\frac{\mathrm{2}{cos}^{\mathrm{5}} \left(\mathrm{3}{x}\right)}{\mathrm{5}}+\frac{{cos}^{\mathrm{7}} \left(\mathrm{3}{x}\right)}{\mathrm{7}}\right]+{c} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18

4)∫((3x+5)/((1−x)(x+3)(x^2 −2)))dx ∫((3x+5)/((1−x)(x+3)(x^2 −2)))dx=∫[(a/(1−x))+(b/(x+3))+(c/(x+(√2) ))+(d/(x−(√2)))]dx =∫(a/(1−x))dx+∫(b/(x+3))dx+∫(c/(x+(√2)))dx+(d/(x−(√2)))dx =(−a)∫(dx/(x−1))+b∫(dx/(x+3))+c∫(dx/(x+(√2)))+d∫(dx/(x−(√2))) =(−a)ln(x−1)+bln(x+3)+cln(x+(√2) )+dln(x−(√2) )+k←(1) now we calculate value of a,b,c and d 3x+5=a(x+3)(x+(√2) )(x−(√2) )+b(1−x)(x+(√2) )(x−(√2) )+c(1−x)(x+3)(x−(√2) )+d(1−x)((x+3)(x+(√2) ) put x=1 ,−3,−(√2) ,(√2) one by one both side 3×1+5=a(1+3)(1+(√2) )(1−(√2) ) 8=−4a a=−2 3×−3+5=b(1+4)(−3+(√2) )(−3−(√2) ) −4=b×5×(9−2) b=((−4)/(35)) 3×−(√2) +5=c(1+(√2) )(−(√2) +3)(−(√2) −(√2) ) −3(√2) +5=c(1+(√2) )(3−(√2) )(−2(√2) ) c=((5−3(√2))/((1+(√2) )((√2)−3)(2(√2)))) 3×(√2) +5=d(1−(√2) )((√2) +3)((√2) +(√2) ) d=((3(√2) +5)/((1−(√2) )(3+(√2) )(2(√2) ))) now put the value in a ,b,c and d in answer (1)

$$\left.\mathrm{4}\right)\int\frac{\mathrm{3}{x}+\mathrm{5}}{\left(\mathrm{1}−{x}\right)\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)}{dx} \\ $$$$\int\frac{\mathrm{3}{x}+\mathrm{5}}{\left(\mathrm{1}−{x}\right)\left({x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{2}\right)}{dx}=\int\left[\frac{{a}}{\mathrm{1}−{x}}+\frac{{b}}{{x}+\mathrm{3}}+\frac{{c}}{{x}+\sqrt{\mathrm{2}}\:}+\frac{{d}}{{x}−\sqrt{\mathrm{2}}}\right]{dx} \\ $$$$=\int\frac{{a}}{\mathrm{1}−{x}}{dx}+\int\frac{{b}}{{x}+\mathrm{3}}{dx}+\int\frac{{c}}{{x}+\sqrt{\mathrm{2}}}{dx}+\frac{{d}}{{x}−\sqrt{\mathrm{2}}}{dx} \\ $$$$=\left(−{a}\right)\int\frac{{dx}}{{x}−\mathrm{1}}+{b}\int\frac{{dx}}{{x}+\mathrm{3}}+{c}\int\frac{{dx}}{{x}+\sqrt{\mathrm{2}}}+{d}\int\frac{{dx}}{{x}−\sqrt{\mathrm{2}}} \\ $$$$=\left(−{a}\right){ln}\left({x}−\mathrm{1}\right)+{bln}\left({x}+\mathrm{3}\right)+{cln}\left({x}+\sqrt{\mathrm{2}}\:\right)+{dln}\left({x}−\sqrt{\mathrm{2}}\:\right)+{k}\leftarrow\left(\mathrm{1}\right) \\ $$$${now}\:{we}\:{calculate}\:{value}\:{of}\:{a},{b},{c}\:{and}\:{d} \\ $$$$\mathrm{3}{x}+\mathrm{5}={a}\left({x}+\mathrm{3}\right)\left({x}+\sqrt{\mathrm{2}}\:\right)\left({x}−\sqrt{\mathrm{2}}\:\right)+{b}\left(\mathrm{1}−{x}\right)\left({x}+\sqrt{\mathrm{2}}\:\right)\left({x}−\sqrt{\mathrm{2}}\:\right)+{c}\left(\mathrm{1}−{x}\right)\left({x}+\mathrm{3}\right)\left({x}−\sqrt{\mathrm{2}}\:\right)+{d}\left(\mathrm{1}−{x}\right)\left(\left({x}+\mathrm{3}\right)\left({x}+\sqrt{\mathrm{2}}\:\right)\right. \\ $$$${put}\:{x}=\mathrm{1}\:,−\mathrm{3},−\sqrt{\mathrm{2}}\:,\sqrt{\mathrm{2}}\:\:{one}\:{by}\:{one}\:\:{both}\:{side} \\ $$$$ \\ $$$$\mathrm{3}×\mathrm{1}+\mathrm{5}={a}\left(\mathrm{1}+\mathrm{3}\right)\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\right) \\ $$$$\mathrm{8}=−\mathrm{4}{a}\:\:\:{a}=−\mathrm{2} \\ $$$$ \\ $$$$\mathrm{3}×−\mathrm{3}+\mathrm{5}={b}\left(\mathrm{1}+\mathrm{4}\right)\left(−\mathrm{3}+\sqrt{\mathrm{2}}\:\right)\left(−\mathrm{3}−\sqrt{\mathrm{2}}\:\right) \\ $$$$−\mathrm{4}={b}×\mathrm{5}×\left(\mathrm{9}−\mathrm{2}\right) \\ $$$${b}=\frac{−\mathrm{4}}{\mathrm{35}} \\ $$$$ \\ $$$$\mathrm{3}×−\sqrt{\mathrm{2}}\:+\mathrm{5}={c}\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)\left(−\sqrt{\mathrm{2}}\:+\mathrm{3}\right)\left(−\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{2}}\:\right) \\ $$$$−\mathrm{3}\sqrt{\mathrm{2}}\:+\mathrm{5}={c}\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)\left(\mathrm{3}−\sqrt{\mathrm{2}}\:\right)\left(−\mathrm{2}\sqrt{\mathrm{2}}\:\right) \\ $$$${c}=\frac{\mathrm{5}−\mathrm{3}\sqrt{\mathrm{2}}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\:\right)\left(\sqrt{\mathrm{2}}−\mathrm{3}\right)\left(\mathrm{2}\sqrt{\mathrm{2}}\right)} \\ $$$$ \\ $$$$\mathrm{3}×\sqrt{\mathrm{2}}\:+\mathrm{5}={d}\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\right)\left(\sqrt{\mathrm{2}}\:+\mathrm{3}\right)\left(\sqrt{\mathrm{2}}\:+\sqrt{\mathrm{2}}\:\right) \\ $$$${d}=\frac{\mathrm{3}\sqrt{\mathrm{2}}\:+\mathrm{5}}{\left(\mathrm{1}−\sqrt{\mathrm{2}}\:\right)\left(\mathrm{3}+\sqrt{\mathrm{2}}\:\right)\left(\mathrm{2}\sqrt{\mathrm{2}}\:\right)} \\ $$$${now}\:{put}\:{the}\:{value}\:{in}\:{a}\:,{b},{c}\:{and}\:{d}\:{in}\:{answer}\:\left(\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$

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