Question

Solve the following

$\int \: \sqrt{ {e}^{x} - 1} \: dx \:$


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Answer 1

Step-by-step explanation:

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Answer 2

$\large\underline{\sf{Solution-}}$

$\rm \: \displaystyle\int\rm \sqrt{{e}^{x} - 1} \:dx \:$

To evaluate this integral, we use method of Substitution.

So, Substitute

$\red{\rm \: \sqrt{{e}^{x} - 1} \: = \: y}$

$\red{\rm \: {e}^{x} - 1 \: = \: {y}^{2} }$

So, on differentiating both sides w. r. t. x, we get

$\red{\rm \: {e}^{x} \: = \:2y \dfrac{dy}{dx} }$

$\red{\rm \: dx \: = \dfrac{2y}{{e}^{x}} \: dy }$

$\red{\rm \: dx \: = \:\dfrac{2y}{ {y}^{2} + 1 } \: dy}$

So, on substituting these values, in above integral, we get

$\rm \: =  \:\displaystyle\int\rm y \times \frac{2y}{ {y}^{2} + 1} \: dy$

$\rm \: =  \:2 \: \displaystyle\int\rm \frac{ {y}^{2} }{ {y}^{2} + 1} \: dy$

$\rm \: =  \:2 \: \displaystyle\int\rm \frac{ {y}^{2} + 1 - 1}{ {y}^{2} + 1} \: dy$

$\rm \: =  \:2 \: \displaystyle\int\rm \bigg[1 - \dfrac{1}{ {y}^{2} + 1} \bigg] \: dy$

$\rm \: =  \:2(y - {tan}^{ - 1}y) + c$

$\rm \: =  \:2( \sqrt{{e}^{x} - 1} - {tan}^{ - 1} \sqrt{{e}^{x} - 1} ) + c$

Hence,

$\rm \: \displaystyle\int\rm \sqrt{{e}^{x} - 1} dx=  \:2( \sqrt{{e}^{x} - 1} - {tan}^{ - 1} \sqrt{{e}^{x} - 1} ) + c$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$