Evalute : 20-12 e^x 3 e^x-4 d x - Vidyadip

Question

Evalute : $\int \frac{20-12 e^x}{3 e^x-4} d x$

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Answer

Let $I =\int \frac{20-12 e^x}{3 e^x-4} d x$
Put, Numerator $= A$ (Denominator $)+ B \left[\frac{d}{d x}\right.$ (Denominator) $]$
$
\begin{aligned}
& \therefore 20-12 e ^{ x }= A \left(3 e ^{ x }-4\right)+ B \left[\frac{d}{d x}\left(3 e ^{ x }-4\right)\right] \\
& \therefore 20-12 e ^{ x }= A \left(3 e ^{ x }-4\right)+ B \left(3 e ^{ x }-0\right) \\
& \therefore 20-12 e ^{ x }=(3 A +3 B ) e ^{ x }-4 A
\end{aligned}
$
Equating the coefficient of ex and constant on both sides, we get
$
\begin{aligned}
& 3 A+3 B=-12 \\
& \text { and }-4 A=20 \\
& \therefore A=-5
\end{aligned}
$
from (1), $3(-5)+3 B=-12$
$
\begin{aligned}
& \therefore 3 B =3 \\
& \therefore B =1 \\
& \therefore 20-12 e ^{ x }=-5\left(3 e ^{ x }-4\right)+\left(3 e ^{ x }\right) \\
& \therefore I=\int\left[\frac{-5\left(3 e^x-4\right)+\left(3 e^x\right)}{3 e^x-4}\right] d x \\
& \quad=\int\left(-5+\frac{3 e^x}{3 e^x-4}\right) d x \\
& \quad=-5 \int 1 d x+\int \frac{3 e^x}{3 e^x-4} d x \\
& \quad=-5 x+\log \left|3 e^x-4\right|+c \\
& \quad \ldots\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right]
\end{aligned}
$

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