∫ex1+ex2+exdx is equal to

# $\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\left(1+{\mathrm{e}}^{\mathrm{x}}\right)\left(2+{\mathrm{e}}^{\mathrm{x}}\right)}\mathrm{dx}$ is equal to

1. A

$\mathrm{log}\left|\frac{2+{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\right|+\mathrm{C}$

2. B

$\mathrm{log}\left|\frac{1+{\mathrm{e}}^{\mathrm{x}}}{1-{\mathrm{e}}^{\mathrm{x}}}\right|+\mathrm{C}$

3. C

$\frac{1}{2}\mathrm{log}\left|\frac{2+{\mathrm{e}}^{\mathrm{x}}}{1+{\mathrm{e}}^{\mathrm{x}}}\right|+\mathrm{C}$

4. D

$\mathrm{log}\left|\frac{1+{\mathrm{e}}^{\mathrm{x}}}{2+{\mathrm{e}}^{\mathrm{x}}}\right|+\mathrm{C}$

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### Solution:

Let,$\mathrm{I}=\int \frac{{\mathrm{e}}^{\mathrm{x}}}{\left(1+{\mathrm{e}}^{\mathrm{x}}\right)\left(2+{\mathrm{e}}^{\mathrm{x}}\right)}\mathrm{dx}$

put,${\mathrm{e}}^{\mathrm{x}}=\mathrm{t}⇒{\mathrm{e}}^{\mathrm{x}}\mathrm{dx}=\mathrm{dt}$

On equating the coefficients of t and constant term on
both sides, we get

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