∫2ex+e−x2dx is equal to - Infinity Learn by Sri Chaitanya

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

Solution:

We have,

$\begin{array}{l} I = \int \frac{2}{{(e^{x} + e^{- x})}^{2}} d x \\ = \int \frac{2 e^{2 x}}{{(e^{2 x} + 1)}^{2}} d x = \int \frac{1}{{(e^{2 x} + 1)}^{2}} d (e^{2 x} + 1) \\ \Rightarrow I = - \frac{1}{e^{2 x} + 1} + C = \frac{- e^{- x}}{e^{x} + e^{- x}} + C \end{array}$

$\int \frac{2}{{(e^{x} + e^{- x})}^{2}} d x$ is equal to

Solution:

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