The value of _0^1 d x e^x+e^ -x is - Vidyadip

MCQ

The value of $\int_0^1 \frac{d x}{e^x+e^{-x}}$ is

A
$\tan ^{-1}\left(\frac{1-e}{1+e}\right)$
✓
$\tan ^{-1}\left(\frac{e-1}{e+1}\right)$
C
$\frac{\pi}{4}$
D
$\tan ^{-1} e+\frac{\pi}{4}$

✓

Answer

Correct option: B.

$\tan ^{-1}\left(\frac{e-1}{e+1}\right)$

(B)
Let $I =\int_0^1 \frac{d x}{ e ^x+ e ^{-x}}=\int_0^1 \frac{ e ^x}{1+ e ^{2 x}} d x$
Put $e ^x= t \Rightarrow e ^x d x= dt$
$\therefore \quad I=\int_1^e \frac{d t}{1+t^2}=\left[\tan ^{-1} t\right]_1^e=\tan ^{-1} e-\tan ^{-1} 1$
$=\tan ^{-1}\left(\frac{ e -1}{ e +1}\right)$
$\ldots\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right]$

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