Evalute : 3 e^x+4 2 e^x-8 d x - Vidyadip

Question

Evalute : $\int \frac{3 e^x+4}{2 e^x-8} d x$

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Answer

Let $I =\int \frac{3 e^x+4}{2 e^x-8} d x$
Put, Numerator $= A$ (Denominator $)+ B \left[\frac{d}{d x}\right.$ (Denominator $\left.)\right]$
$
\begin{aligned}
& \therefore 3 e ^{ x }+4= A \left(2 e ^{ x }-8\right)+ B \left[\frac{d}{d x}\left(2 e ^{ x }-8\right)\right] \\
& \therefore 3 e ^{ x }+4= A \left(2 e ^{ x }-8\right)+ B \left(2 e ^{ x }-0\right) \\
& \therefore 3 e ^{ x }+4=(2 A +2 B ) e ^{ x }-8 A
\end{aligned}
$
Equating the coefficient of ex and constant on both sides, we get
$
\begin{aligned}
& 2 A +2 B =3 \\
& \text { and }-8 A =4
\end{aligned}
$
$
\therefore A =-\frac{1}{2}
$
$\therefore$ from $(1), 2\left(-\frac{1}{2}\right)+2 B=3$
$
\therefore 2 B =4
$
$
\therefore B =2
$
$
\therefore 3 e^x+4=-\frac{1}{2}\left(2 e^x-8\right)+2\left(2 e^x\right)
$
$\therefore I=\int\left[\frac{-\frac{1}{2}\left(2 e^x-8\right)+2\left(2 e^x\right)}{2 e^x-8}\right] d x$
$
=\int\left[-\frac{1}{2}+\frac{2\left(2 e^x\right)}{2 e^x-8}\right] d x
$
$
=-\frac{1}{2} \int 1 d x+2 \int \frac{2 e^x}{2 e^x-8} d x
$
$
=-\frac{1}{2} x+2 \log \left|2 e^x-8\right|+c
$
$
\ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]
$

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