We have, I=∫e^xdx( 1 + e^2x) Let t=e^x dtdx=e^x dt=e^x dx Therefore, I=∫dt1+t^2 I=tan^ 1 t+C On putting the value of t , we get ...

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Area of a Triangle Given Its Vertices

∫exdx 1 + e2x

Question

$\int \frac{e^{x} d x}{(1 + e^{2 x})}$

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Solution

We have,
$I = \int \frac{e^{x} d x}{(1 + e^{2 x})}$

Let
$t = e^{x}$

$\frac{d t}{d x} = e^{x}$

$d t = e^{x} d x$

Therefore,
$I = \int \frac{d t}{1 + t^{2}}$

$I = {tan}^{- 1} t + C$

On putting the value of $t$ , we get
$I = {tan}^{- 1} e^{x} + C$

Hence, this is the answer.

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