int frac{dx}{(x^2 + 1)(x^2 + 4)} = | vedclass

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(C) Let $I = \int \frac{dx}{(x^2 + 1)(x^2 + 4)}$.Using partial fractions,let $t = x^2$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$.$1

Vedclass · Accepted Answer

$\frac{1}{3}	an^{-1}x - \frac{1}{6}	an^{-1}\frac{x}{2} + c$

Vedclass · Answer

(C) Let $I = \int \frac{dx}{(x^2 + 1)(x^2 + 4)}$.Using partial fractions,let $t = x^2$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$.$1 = A(t+4) + B(t+1)$.For $t = -1$,$1 = 3A \Rightarrow A = \frac{1}{3}$.For $t = -4$,$1 = -3B \Rightarrow B = -\frac{1}{3}$.So,$\frac{1}{(x^2+1)(x^2+4)} = \frac{1}{3} \left( \frac{1}{x^2+1} - \frac{1}{x^2+4} ight)$.Integrating both sides:$I = \frac{1}{3} \int \frac{dx}{x^2+1} - \frac{1}{3} \int \frac{dx}{x^2+2^2}$.Using the formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + c$:$I = \frac{1}{3} 	an^{-1}x - \frac{1}{3} \left( \frac{1}{2} 	an^{-1}\frac{x}{2} ight) + c$.$I = \frac{1}{3} 	an^{-1}x - \frac{1}{6} 	an^{-1}\frac{x}{2} + c$.

$\int \frac{dx}{(x^2 + 1)(x^2 + 4)} = $

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