Question

Solve the following

$\int \: \frac{dx}{a + {be}^{x} }$
​

Answer 1

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

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 :-

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

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

Answer 2

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

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 :-

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