First slide

Methods of integration

Question

$\int \sqrt{e^{x} - 1} d x =$

Easy

A

$2 [\sqrt{e^{x} - 1} - \tan^{- 1} \sqrt{e^{x} - 1}] + c$
B

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

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

$2 [\sqrt{e^{x} - 1} + \tan^{- 1} \sqrt{e^{x} - 1}] + c$

Solution

$\begin{array}{l} L e t I = \int \sqrt{e^{x} - 1} d x \\ P u t e^{x} - 1 = t^{2} ⇒ e^{x} d x = 2 t d t \\ \Rightarrow d x = \frac{2 t}{1 + t^{2}} d t \\ I = \int t . \frac{2 t}{1 + t^{2}} d t \\ = \int \frac{2 t^{2}}{1 + t^{2}} d t \\ = \int \frac{2 t^{2} + 2 - 2}{1 + t^{2}} d t \\ = 2 \int \frac{t^{2} + 1}{1 + t^{2}} d t - 2 \int \frac{1}{1 + t^{2}} d t \\ = 2 t - 2 \tan^{- 1} t + c \\ = 2 [\sqrt{e^{x} - 1} - \tan^{- 1} \sqrt{e^{x} - 1}] + c \end{array}$

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