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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 16 Maths Textbook Solution.

Answers (1)

Answer:

$x-\log \left|e^{x}+1\right|+c$

Given:

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

Hint:

Do integration by separation.

Solution:

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

$=\int \frac{\left(e^{x}+1\right)\left(-e^{x}\right)}{e^{x}+1} d x$

$=\int d x-\int \frac{e^{x}}{e^{x}+1} d x$

$now, let e^{x}+1=u$

Differentiate it with respect to x

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

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

now,

$I=\int d x-\int \frac{1}{u} d u$

$=x-\log |u|+c$

$=x-\log \left|e^{x}+1\right|+c \quad\left(\because u=e^{x}+1\right)$

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