Evaluate : x ^ -1 x d x - Vidyadip

Question

Evaluate : $\int x \cdot \tan ^{-1} x \cdot d x$

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Answer

$
\begin{aligned}
\text { I } & =\int\left(\tan ^{-1} x \cdot\right) x \cdot d x \quad \ldots . \text { by LIATE } \\
& =\tan ^{-1} x \cdot \int x \cdot d x-\int \frac{d}{d x} \cdot \tan ^{-1} x \cdot \int x \cdot d x \cdot d x \\
& =\tan ^{-1} x \cdot \frac{x^2}{2}-\int \frac{1}{1+x^2} \cdot \frac{x^2}{2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} \int \frac{x^2}{1+x^2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} \int \frac{1+x^2-1}{1+x^2} \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} \int\left[1-\frac{1}{1+x^2}\right] \cdot d x \\
& =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2}\left[x-\tan ^{-1} x\right]+c \\
\therefore \quad \int x \cdot \tan ^{-1} x \cdot d x & =\frac{1}{2} x^2 \cdot \tan ^{-1} x-\frac{1}{2} x+\frac{1}{2} \tan ^{-1} x+c
\end{aligned}
$

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