1- e x -12 dx is equal toA- e x -1-1- e x -1- CB- -log e x -1-1- e x -1- CC- -x-1-ex-1-logex-1-CD- x-logex-1-1-ex-1-C

The correct option is C x 1(e^x 1) log(e^x 1)+C∫1(e^x 1)^2 dx=∫e^x (e^x 1)(e^x 1)^2 dx =∫e^x(e^x 1)^2 ∫1(e^x 1) dx =∫e^x(e^x 1)^2 dx ∫e^x (e^x 1)(e^x 1) dx =∫ ...

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Byju's Answer

Standard XII

Mathematics

Integration of Piecewise Continuous Functions

1/ e x -12 dx...

Question

$\int \frac{1}{(e^{x} - 1)^{2}} d x$ is equal to

(e^{x} - 1) - \frac{1}{(e^{x} - 1)} + C

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- log (e^{x} - 1) - \frac{1}{(e^{x} - 1)} + C

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- x - \frac{1}{(e^{x} - 1)} - log (e^{x} - 1) + C

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x + log (e^{x} - 1) - \frac{1}{(e^{x} - 1)} + C

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Solution

The correct option is C $- x - \frac{1}{(e^{x} - 1)} - log (e^{x} - 1) + C$
$\int \frac{1}{(e^{x} - 1)^{2}} d x = \int \frac{e^{x} - (e^{x} - 1)}{(e^{x} - 1)^{2}} d x = \int \frac{e^{x}}{(e^{x} - 1)^{2}} - \int \frac{1}{(e^{x} - 1)} d x = \int \frac{e^{x}}{(e^{x} - 1)^{2}} d x - \int \frac{e^{x} - (e^{x} - 1)}{(e^{x} - 1)} d x = \int \frac{e^{x}}{(e^{x} - 1)^{2}} d x - \int \frac{e^{x}}{(e^{x} - 1)} d x - \int 1 d x$
Let $t = e^{x} - 1 \Rightarrow d t = e^{x} d x$
$\int (\frac{1}{t^{2}} - \frac{1}{t}) d t - \int 1 d x = - \frac{1}{t} - log t - x + C = - x - \frac{1}{(e^{x} - 1)} - log (e^{x} - 1) + C$

Suggest Corrections

∫1(ex−1)2 dx is equal to

$\int \frac{1}{(e^{x} - 1)^{2}} d x$ is equal to