Math, asked by khushi15686, 2 months ago

Solve the following

 \int \:  \frac{dx}{a +  {be}^{x} }

Answers

Answered by mathdude500
7

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

Given integral is

\rm \: \displaystyle\int\rm  \frac{dx}{a \:  +  \: b \:  {e}^{x} }  \\

can be rewritten as

\rm \: =  \:  \displaystyle\int\rm  \frac{1}{a \:  +  \: \dfrac{b}{{e}^{ - x}}  } \: dx \:   \\

\rm \: =  \:  \displaystyle\int\rm  \frac{1}{\dfrac{a{e}^{ - x} + b}{{e}^{ - x}}  } \: dx \:   \\

\rm \: =  \:\displaystyle\int\rm  \frac{{e}^{ - x}}{a{e}^{ - x} + b}  \: dx \\

Now, we use substitution method to solve this integral.

So, Substitute

\rm \: a{e}^{ - x} + b = y \\

\rm \: \dfrac{d}{dx}( a{e}^{ - x} + b) = \dfrac{d}{dx}y \\

\rm \: -  a{e}^{ - x} \:  =  \: \dfrac{dy}{dx} \\

\rm \: {e}^{ - x} \: dx \:  =  \:   -  \: \dfrac{dy}{a}  \\

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

\rm \: =  \: - \dfrac{1}{a}\displaystyle\int\rm  \frac{dy}{y}  \\

\rm \: =  \: - \dfrac{1}{a} \: log |y| + c   \\

\rm \: =  \: - \dfrac{1}{a} \: log \bigg|a{e}^{ - x} + b\bigg| + c   \\

Hence,

\boxed{\sf{  \: \: \rm \:\displaystyle\int\rm  \frac{dx}{a + b{e}^{x}}  =  \: -  \: \dfrac{1}{a} \: log \bigg|a{e}^{ - x} + b\bigg| + c  \:  \: }}  \\

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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}

\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}

Answered by nihasrajgone2005
1

\huge\red{A}\pink{N}\orange{S} \green{W}\blue{E}\gray{R} =

Given integral is

\begin{gathered}\rm \: \displaystyle\int\rm \frac{dx}{a \: + \: b \: {e}^{x} } \\ \end{gathered}

can be rewritten as

\begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{1}{a \: + \: \dfrac{b}{{e}^{ - x}} } \: dx \: \\ \end{gathered}  \\ \begin{gathered}\rm \: = \: \displaystyle\int\rm \frac{1}{\dfrac{a{e}^{ - x} + b}{{e}^{ - x}} } \: dx \: \\ \end{gathered}  \\ \begin{gathered}\rm \: = \:\displaystyle\int\rm \frac{{e}^{ - x}}{a{e}^{ - x} + b} \: dx \\ \end{gathered}

Now, we use substitution method to solve this integral.

So, Substitute

\begin{gathered}\rm \: a{e}^{ - x} + b = y \\ \end{gathered}  \\ \begin{gathered}\rm \: \dfrac{d}{dx}( a{e}^{ - x} + b) = \dfrac{d}{dx}y \\ \end{gathered}  \\ \begin{gathered}\rm \: - a{e}^{ - x} \: = \: \dfrac{dy}{dx} \\ \end{gathered}  \\ \begin{gathered}\rm \: {e}^{ - x} \: dx \: = \: - \: \dfrac{dy}{a} \\ \end{gathered}

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

\begin{gathered}\rm \: = \: - \dfrac{1}{a}\displaystyle\int\rm \frac{dy}{y} \\ \end{gathered} \\ \begin{gathered}\rm \: = \: - \dfrac{1}{a} \: log |y| + c \\ \end{gathered}  \\ \begin{gathered}\rm \: = \: - \dfrac{1}{a} \: log \bigg|a{e}^{ - x} + b\bigg| + c \\ \end{gathered} Hence

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Additional Information :-

\begin{gathered}\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}\end{gathered}

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