Math, asked by chandavedic, 17 days ago

Integrate the following function

 \frac{1}{ \sqrt{{e}^{x} - 4 }}

Answers

Answered by mathdude500
6

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

Given integral is

\rm \: \displaystyle\int\rm  \frac{dx}{ \sqrt{ {e}^{x}  - 4} }  \\

To evaluate this integral, we use method of Substitution.

So, Substitute

\rm \:  \sqrt{{e}^{x} - 4}  = y \\

\rm \: {e}^{x} - 4 =  {y}^{2}  \\

\rm \: {e}^{x} =  {y}^{2} + 4  \\

On differentiating both sides w. r. t. x, we get

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

\rm \: ( {y}^{2} + 4)  \: dx =  2y \: dy  \\

\rm \: dx \:  =  \: \dfrac{2y}{ {y}^{2}  + 4}  \: dy \\

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

\rm \:  = \displaystyle\int\rm  \frac{1}{y} \times  \frac{2y}{ {y}^{2}  + 4}  \: dy \\

\rm \:  =2 \displaystyle\int\rm   \frac{1}{ {y}^{2}  + 4}  \: dy \\

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

\rm \:  = 2 \times \dfrac{1}{2}  {tan}^{ - 1}\dfrac{y}{2}  + c \\

\rm \:  = {tan}^{ - 1}\dfrac{y}{2}  + c \\

\rm \:  = {tan}^{ - 1}\dfrac{ \sqrt{{e}^{x} - 4}}{2}  + c \\

Hence,

\rm\implies \:\boxed{ \rm{ \:\rm \: \displaystyle\int\rm  \frac{dx}{ \sqrt{{e}^{x} - 4} }  = {tan}^{ - 1}\dfrac{ \sqrt{{e}^{x} - 4}}{2}  + c \: }} \\

\rule{190pt}{2pt}

Additional information :-

\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{\displaystyle\int\sf  \frac{dx}{ {x}^{2}  +  {a}^{2} }  =  \dfrac{1}{a} {tan}^{ - 1} \dfrac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf  \frac{dx}{ \sqrt{ {x}^{2}  -  {a}^{2} } }  = log |x +  \sqrt{ {x}^{2}  -  {a}^{2} } | + c  }\\ \\ \bigstar \: \bf{\displaystyle\int\sf  \frac{dx}{ \sqrt{ {a}^{2}  -  {x}^{2} } }  =  {sin}^{ - 1}  \frac{x}{a} + c }\\ \\ \bigstar \: \bf{\displaystyle\int\sf  \frac{dx}{ \sqrt{ {x}^{2}  +  {a}^{2} } } = log |x +  \sqrt{ {x}^{2} +  {a}^{2}} | + c}\\ \\  \end{array} }}\end{gathered}\end{gathered}\end{gathered}

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