First slide

Methods of integration

Question

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

Easy

A

$\frac{e^{\tan^{- 1} x}}{1 + x^{2}} + C$
B

$\frac{x e^{\tan^{- 1} x}}{1 + x^{2}} + C$
C

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

none of these

Solution

Putting $\tan^{- 1} x = u, we have \frac{d x}{1 + x^{2}} = d u$
$\begin{array}{l} \int e^{\tan^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x = \int e^{u} (1 + \tan u + \tan^{2} u) d u \\ = \int e^{u} (\sec^{2} u + \tan u) d u \\ = \tan u e^{u} + C = x e^{\tan^{- 1} x} + C . \end{array}$
$(using \int e^{x} (f (x) + f^{'} (x) d x = e^{x} f (x) + C)$

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