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

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(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)(

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

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(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 = $

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