Question
Prove that $\int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c$
✓
Answer
$\begin{aligned} & \int \frac{1}{a^2-x^2} d x=\int \frac{1}{(a-x)(a+x)} d x \\ = & \frac{1}{2 a} \int \frac{(a-x)+(a+x)}{(a-x)(a+x)} d x \\ = & \frac{1}{2 a} \int\left(\frac{1}{a+x}+\frac{1}{a-x}\right) d x \\ = & \frac{1}{2 a}\left[\int \frac{1}{a+x} d x+\int \frac{1}{a-x} d x\right] \\ = & \frac{1}{2 a}\left[\log |a+x|+\frac{\log |a-x|}{-1}\right]+C=\frac{1}{2 a}[\log |a+x|-\log |a-x|]+C \\ = & \frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\end{aligned}$
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