Question

Evaluate the following integration
$\int \frac{dx}{a + b {e}^{x} }$
​

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 :-

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