int {frac{{{x^2}}}{{left( {{x^2} + 1} right)l | vedclass

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(B) Let $I = \int \frac{x^{2}}{\left(x^{2}+1\right)\left(x^{2}+4\right)} dx$. Substitute $x^{2} = y$. Then,we can write the integrand as $\frac{y}{(y+

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \frac{x^{2}}{\left(x^{2}+1ight)\left(x^{2}+4ight)} dx$. Substitute $x^{2} = y$. Then,we can write the integrand as $\frac{y}{(y+1)(y+4)}$. Using partial fractions,$\frac{y}{(y+1)(y+4)} = \frac{A}{y+1} + \frac{B}{y+4}$. $y = A(y+4) + B(y+1)$. For $y = -1$,$-1 = A(3) \implies A = -\frac{1}{3}$. For $y = -4$,$-4 = B(-3) \implies B = \frac{4}{3}$. Thus,$I = \int \left( \frac{-1/3}{x^{2}+1} + \frac{4/3}{x^{2}+4} ight) dx$. $I = -\frac{1}{3} \int \frac{1}{x^{2}+1} dx + \frac{4}{3} \int \frac{1}{x^{2}+2^{2}} dx$. Using the formula $\int \frac{1}{x^{2}+a^{2}} dx = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$,we get: $I = -\frac{1}{3} 	an^{-1}(x) + \frac{4}{3} \cdot \frac{1}{2} 	an^{-1}(\frac{x}{2}) + C$. $I = -\frac{1}{3} 	an^{-1}(x) + \frac{2}{3} 	an^{-1}(\frac{x}{2}) + C$.

$\int {\frac{{{x^2}}}{{\left( {{x^2} + 1} \right)\left( {{x^2} + 4} \right)}}\,} dx$ is equal to

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