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

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(D) To solve the integral $I = \int \frac{dx}{(x^2+1)(x^2+4)}$, we use partial fractions. Let $x^2 = t$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} +

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$\frac{1}{3} 	an^{-1} x - \frac{1}{6} 	an^{-1}(\frac{x}{2}) + C$

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(D) To solve the integral $I = \int \frac{dx}{(x^2+1)(x^2+4)}$, we use partial fractions. Let $x^2 = t$. Then $\frac{1}{(t+1)(t+4)} = \frac{A}{t+1} + \frac{B}{t+4}$.By partial fraction decomposition, $1 = A(t+4) + B(t+1)$.Setting $t = -1$, we get $1 = 3A \implies A = \frac{1}{3}$.Setting $t = -4$, we get $1 = -3B \implies B = -\frac{1}{3}$.Thus, $\frac{1}{(x^2+1)(x^2+4)} = \frac{1}{3} \left( \frac{1}{x^2+1} - \frac{1}{x^2+4} ight)$.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 standard formula $\int \frac{dx}{x^2+a^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$, we get:$I = \frac{1}{3} 	an^{-1} x - \frac{1}{3} \cdot \frac{1}{2} 	an^{-1}(\frac{x}{2}) + C$.$I = \frac{1}{3} 	an^{-1} x - \frac{1}{6} 	an^{-1}(\frac{x}{2}) + C$.

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

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