First slide

Methods of integration

Question

$\int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} dx$ is equal to

Easy

A

$- e^{\tan^{- 1} x} + C$
B

$e^{\tan^{- 1} x} + C$
C

$\tan^{- 1} x + C$
D

$- \tan^{- 1} x + C$

Solution

Here, the given integrand is of the form ,e^f(x), f ' (x),so we use substitution method to integrate it.
$\begin{array}{l} \int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} dx \\ Let \tan^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} = \frac{dt}{dx} \\ \Rightarrow dx = (1 + x^{2}) dt \\ ∴ \int \frac{e^{\tan^{- 1} x}}{1 + x^{2}} dx = \int \frac{e^{t}}{1 + x^{2}} (1 + x^{2}) dt \\ = \int e^{t} dt = e^{t} + C = e^{\tan^{- 1} x} + C \end{array}$

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