2 (e^ x +e^ -x )^2 dx= - Vidyadip

MCQ

$\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}=$

✓
$\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
B
$-\frac{1}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
C
$\frac{-1}{(\text{e}^{\text{x}}+1)^2}+\text{C}$
D
$\frac{1}{\text{e}^{\text{x}}-\text{e}^{-\text{x}}}+\text{C}$

✓

Answer

Correct option: A.

$\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

$\text{I}=\int\frac{2}{(\text{e}^{\text{x}}+\text{e}^{-\text{x}})^2}\text{ dx}$
$\text{I}=\int\frac{2\text{e}^{2\text{x}}}{(\text{e}^{2\text{x}}+1)^2}\text{ dx}$
Put $\text{t}=\text{e}^{2\text{x}}+1$
$\text{dt}=2\text{e}^{2\text{x}}\text{ dx}$
$\text{I}=\int\frac{\text{dt}}{\text{t}^2}$
$\text{I}=\frac{-1}{\text{t}}+\text{C}$
$\text{I}=\frac{-1}{\text{e}^{2\text{x}}+1}+\text{C}$
$\text{I}=\frac{-\frac{1}{\text{e}^{\text{x}}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$
$\text{I}=\frac{-\text{e}^{-\text{x}}}{\text{e}^{\text{x}}+\text{e}^{-\text{x}}}+\text{C}$

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