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Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 17 Maths Textbook Solution.

Answers (1)

Answer:

$2 \log \left|e^{x}+1\right|-x+c$

Given:

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

Hint:

Use partial fraction method.

Solution:

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

$\operatorname{let} e^{\mathrm{x}}=u$

$e^{x} d x=d u$ $(differentiate with respect to x)$

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

$now,$

$I=\int \frac{u-1}{u+1} \times \frac{d u}{u} \quad(p u t u \& d u)$

use partial fraction

$u-1=A(1+u)+B u$

$u-1=A+A u+B u$

$A=-1 \& B+A=1 \quad \therefore B=2 \quad (on comparing)$

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

$=-\int \frac{1}{u}+\frac{2}{u+1} d u$

$=-\log |\mathrm{u}|+2 \log |\mathrm{u}+1|+\mathrm{c}$

$=-\log \left|\mathrm{e}^{\mathrm{x}}\right|+2 \log \left|\mathrm{e}^{\mathrm{x}}+1\right|+\mathrm{c}$

$=-2 \log \left|\mathrm{e}^{x}+1\right|-x+c \ldots(\operatorname{loge}=1)$

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