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

Answers (1)

Answer:

$\tan ^{-1}\left(e^{x}\right)+c$

Given:

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

Hint:

Use substitution method.

Solution:

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

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

$=\frac{e^{x}}{\left(e^{x}\right)^{2}+1} d x$

$let u=e^{x}$

$d u=e^{x} d x$ $(on differentiating)$

$now,$

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

$=\tan ^{-1} u+c$

$=\tan ^{-1}\left(\mathrm{e}^{\mathrm{x}}\right)+\mathrm{c}$

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