First slide

Introduction to integration

Question

$\int e^{\tan^{- 1} x} (1 + x + x^{2}) \cdot d (\cot^{- 1} x) is equal to$

Moderate

A

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

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

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

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

Solution

$\begin{array}{l} I = \int e^{\tan^{- 1} x} (1 + x + x^{2}) \cdot (- \frac{1}{1 + x^{2}}) d x \\ = - \int e^{t {an}^{- 1} x} (1 + \frac{x}{1 + x^{2}}) d x \\ = - \int e^{ta n^{- 1} x} d x - \{\int x \cdot \frac{e^{\tan^{- 1} x}}{1 + x^{2}} d x\} \\ (i n t e g r a t i o n b y p a r t s) \\ = - \int e^{\tan^{- 1} x} d x - x \cdot e^{\tan^{- 1} x} + \int e^{\tan^{- 1} x} d x + C \\ = - x \cdot e^{\tan^{- 1} x} + C \end{array}$

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