# Chapter 8 - Techniques of Integration - Chapter Review Exercises - Page 460: 52

$$\ln x-\frac{1}{2} \ln \left|1+x^{2}\right|-\frac{\tan ^{-1} x}{x}+C$$

#### Work Step by Step

Given $$\int x^{-2} \tan ^{-1} x d x$$ Let \begin{align*} u&= \tan ^{-1} x\ \ \ \ \ \ dv=x^{-2} d x\\ du&=\frac{1}{1+x^2} dx\ \ \ \ \ \ v=-x^{-1} \end{align*} Then \begin{align*} \int x^{-2} \tan ^{-1} x d x&= -\frac{\tan ^{-1} x}{x}+\int \frac{1}{x\left(1+x^{2}\right)} d x \end{align*} Since \begin{aligned} \frac{1}{x\left(1+x^{2}\right)} &=\frac{A}{x}+\frac{B x+C}{1+x^{2}} \\ 1 &=A\left(1+x^{2}\right)+(B x+C) x \end{aligned} At $x=0$, $A= 1$ and by comparing $C= 0,\ \ B= -1$, we have: \begin{aligned} \int x^{-2} \tan ^{-1} x d x &=-\frac{\tan ^{-1} x}{x}+\int \frac{1}{x\left(1+x^{2}\right)} d x \\ &=-\frac{\tan ^{-1} x}{x}+\int \frac{1}{x} d x-\int \frac{x}{\left(1+x^{2}\right)} d x \\ &=\ln x-\frac{1}{2} \ln \left|1+x^{2}\right|-\frac{\tan ^{-1} x}{x}+C \end{aligned}

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