int frac{1}{x^2} log(x^2 + a^2) , dx = | vedclass

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(B) Let $I = \int x^{-2} \log(x^2 + a^2) \, dx$. Using integration by parts,let $u = \log(x^2 + a^2)$ and $dv = x^{-2} \, dx$. Then $du = \frac{2x}{x^

Vedclass · Accepted Answer

$-\frac{1}{x} \log(x^2 + a^2) + \frac{2}{a} 	an^{-1} \frac{x}{a} + c$

Vedclass · Answer

(B) Let $I = \int x^{-2} \log(x^2 + a^2) \, dx$. Using integration by parts,let $u = \log(x^2 + a^2)$ and $dv = x^{-2} \, dx$. Then $du = \frac{2x}{x^2 + a^2} \, dx$ and $v = -\frac{1}{x}$. Applying the formula $\int u \, dv = uv - \int v \, du$: $I = -\frac{1}{x} \log(x^2 + a^2) - \int \left(-\frac{1}{x}ight) \left(\frac{2x}{x^2 + a^2}ight) \, dx$ $I = -\frac{1}{x} \log(x^2 + a^2) + \int \frac{2}{x^2 + a^2} \, dx$ $I = -\frac{1}{x} \log(x^2 + a^2) + 2 \left(\frac{1}{a} 	an^{-1} \frac{x}{a}ight) + c$ $I = -\frac{1}{x} \log(x^2 + a^2) + \frac{2}{a} 	an^{-1} \frac{x}{a} + c$.

$\int \frac{1}{x^2} \log(x^2 + a^2) \, dx = $

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