int {frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int {\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}} \,dx$. Substitute $t = {x^2}$,then the partial fraction decomposition of $\frac{t}{(t+2)(

Vedclass · Accepted Answer

$ - \sqrt 2 {	an ^{ - 1}}\frac{x}{{\sqrt 2 }} + \sqrt 3 {	an ^{ - 1}}\frac{x}{{\sqrt 3 }} + c$

Vedclass · Answer

(B) Let $I = \int {\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}} \,dx$. Substitute $t = {x^2}$,then the partial fraction decomposition of $\frac{t}{(t+2)(t+3)}$ is $\frac{A}{t+2} + \frac{B}{t+3}$. $t = A(t+3) + B(t+2)$. For $t = -2$,$-2 = A(1) \implies A = -2$. For $t = -3$,$-3 = B(-1) \implies B = 3$. So,$\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}} = \frac{3}{{{x^2} + 3}} - \frac{2}{{{x^2} + 2}}$. Integrating both sides,$I = \int {\frac{3}{{{x^2} + 3}}} \,dx - \int {\frac{2}{{{x^2} + 2}}} \,dx$. Using the formula $\int {\frac{1}{{{x^2} + {a^2}}}} \,dx = \frac{1}{a}{	an ^{ - 1}}\frac{x}{a} + c$,we get: $I = 3 \cdot \frac{1}{{\sqrt 3 }}{	an ^{ - 1}}\frac{x}{{\sqrt 3 }} - 2 \cdot \frac{1}{{\sqrt 2 }}{	an ^{ - 1}}\frac{x}{{\sqrt 2 }} + c$. $I = \sqrt 3 {	an ^{ - 1}}\frac{x}{{\sqrt 3 }} - \sqrt 2 {	an ^{ - 1}}\frac{x}{{\sqrt 2 }} + c$.

$\int {\frac{{{x^2}}}{{({x^2} + 2)({x^2} + 3)}}} \,dx = $

Similar Questions

If $\int \frac{2 x^2-3}{\left(x^2-4\right)\left(x^2+1\right)} d x=A \tan^{-1} x+B \log (x-2)+C \log (x+2)$,then $6 A+7 B-5 C=$

$\int \frac{1}{(x-2)(x^2+1)} dx=$

If $\int \frac{2 x^{2}+3}{\left(x^{2}-1\right)\left(x^{2}+4\right)} d x=a \log \left|\frac{x-1}{x+1}\right|+b \tan ^{-1}\left(\frac{x}{2}\right)+C$,then

Evaluate the integral: $\int \frac{\sin x}{\sin 4x} dx$

$\int \frac{dx}{x(x^{2}+1)}$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module