_0^1 ^ -1 x d x= - Vidyadip

MCQ

$\int_0^1 \tan ^{-1} x d x=$

✓
$\frac{\pi}{4}-\frac{1}{2} \log 2$
B
$\pi-\frac{1}{2} \log 2$
C
$\frac{\pi}{4}-\log 2$
D
$\pi-\log 2$

✓

Answer

Correct option: A.

$\frac{\pi}{4}-\frac{1}{2} \log 2$

(A)
$\int_0^1 \tan ^{-1} x d x=\left[\left(\tan ^{-1} x\right) \cdot x\right]_0^1-\int_0^1 \frac{1}{1+x^2} \cdot x d x$
$\begin{array}{l}=\left[x \tan ^{-1} x-\frac{1}{2} \log \left|1+x^2\right|\right]_0^1 \\ =\frac{\pi}{4}-\frac{1}{2} \log 2\end{array}$

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