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Question Number 42792 by maxmathsup by imad last updated on 02/Sep/18
find ∫_0 ^1 ((1+x^2 )/(1+x^3 ))dx
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} }{dx}\: \\ $$
Commented by behi83417@gmail.com last updated on 03/Sep/18
I=(1/3)ln(1+x^3 )+A.ln(1+x)+B.ln((x−a)/(x−b))+c
$${I}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)+{A}.{ln}\left(\mathrm{1}+{x}\right)+{B}.{ln}\frac{{x}−{a}}{{x}−{b}}+{c} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 04/Sep/18
∫_0 ^1 (dx/(1+x^3 ))+(1/3)∫_0 ^1 ((3x^2 )/(1+x^3 )) ∫_0 ^1 (dx/((1+x)(1−x+x^2 )))+(1/3)∫_0 ^1 ((d(1+x^3 ))/(1+x^3 )) I_2 =(1/3)∫_0 ^1 ((d(1+x^3 ))/(1+x^3 ))=(1/3)∣ln(1+x^3 )∣_0 ^1 =(1/3)ln2 now (1/((1+x)(1−x+x^2 )))=(a/(1+x))+((bx+c)/(x^2 −x+1)) 1=a(x^2 −x+1)+(bx+c)(1+x) 1=ax^2 −ax+a+bx+bx^2 +c+cx contd 1=x^2 (a+b)+x(−a+b+c)+(a+c) a+b=0 −a+b+c=0 a+c=1 −a−a+c=0 c=2a a+2a=1 a=(1/3) b=((−1)/3) c=(2/3) I_1 =∫_0 ^1 {(a/(1+x))+((bx+c)/(x^2 −x+1)) }dx =(1/3)∫_(0 ) ^1 (dx/(1+x))+(1/3)∫_0 ^1 ((−x+2)/(x^2 −x+1))dx =(1/3)∫_0 ^1 (dx/(1+x))−(1/6)∫_0 ^1 ((2x−1−3)/(x^2 −x+1))dx =(1/3)ln2−(1/6)∫_0 ^1 ((d(x^2 −x+1))/(x^2 −x+1))+(1/2)∫_0 ^1 (dx/((x^2 −2.x.(1/2)+(1/4)+1−(1/4)))) =(1/3)ln2−(1/6)∣ln(x^2 −x+1)∣_0 ^1 +(1/2)∫_0 ^1 (dx/((x−(1/2))^2 +(((√3)/2))^2 )) =(1/3)ln2−(1/6)×0+(1/2)×(2/( (√3)))∣tan^(−1) (((x−(1/2))/((√3)/2)))∣_0 ^1 =(1/3)ln2+(1/( (√3))){tan^(−1) (1/( (√3)))−tan^(−1) (−(1/( (√3))))} =(1/3)ln2+(2/( (√3)))×(Π/6)=(1/3)ln2+(Π/(3(√3))) clmplete ans is =(2/3)ln2+(Π/(3(√3)))
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}+\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$$${I}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)}{\mathrm{1}+{x}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{3}}\mid{ln}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)\mid_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2} \\ $$$${now} \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}+{x}\right)\left(\mathrm{1}−{x}+{x}^{\mathrm{2}} \right)}=\frac{{a}}{\mathrm{1}+{x}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}} \\ $$$$\mathrm{1}={a}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)+\left({bx}+{c}\right)\left(\mathrm{1}+{x}\right) \\ $$$$\mathrm{1}={ax}^{\mathrm{2}} −{ax}+{a}+{bx}+{bx}^{\mathrm{2}} +{c}+{cx} \\ $$$${contd} \\ $$$$\mathrm{1}={x}^{\mathrm{2}} \left({a}+{b}\right)+{x}\left(−{a}+{b}+{c}\right)+\left({a}+{c}\right) \\ $$$${a}+{b}=\mathrm{0}\:\:\:−{a}+{b}+{c}=\mathrm{0}\:\:\:\:{a}+{c}=\mathrm{1} \\ $$$$−{a}−{a}+{c}=\mathrm{0}\:\:\:{c}=\mathrm{2}{a}\:\:\:\:{a}+\mathrm{2}{a}=\mathrm{1} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{3}}\:\:\:{b}=\frac{−\mathrm{1}}{\mathrm{3}}\:\:\:{c}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\frac{{a}}{\mathrm{1}+{x}}+\frac{{bx}+{c}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}\:\right\}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{−{x}+\mathrm{2}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\mathrm{1}+{x}}−\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{2}{x}−\mathrm{1}−\mathrm{3}}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{d}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{{x}^{\mathrm{2}} −{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}^{\mathrm{2}} −\mathrm{2}.{x}.\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}}\mid{ln}\left({x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{6}}×\mathrm{0}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\mid{tan}^{−\mathrm{1}} \left(\frac{{x}−\frac{\mathrm{1}}{\mathrm{2}}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\right)\mid_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\left\{{tan}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}−{tan}^{−\mathrm{1}} \left(−\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2}+\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}×\frac{\Pi}{\mathrm{6}}=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mathrm{2}+\frac{\Pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${clmplete}\:{ans}\:{is}\:=\frac{\mathrm{2}}{\mathrm{3}}{ln}\mathrm{2}+\frac{\Pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$

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