Evalute : 2 e^x-3 4 e^x+1 d x - Vidyadip

Question

Evalute : $\int \frac{2 e^x-3}{4 e^x+1} d x$

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Answer

Let $I =\int \frac{2 e ^{ x }-3}{4 e ^{ x }+1} dx$
Let $2 e ^{ x }-3= A \left(4 e ^{ x }+1\right)+ B \frac{ d }{ dx }\left(4 e ^{ x }+1\right)$
$
\therefore 2 e ^{ x }-3=(4 A +4 B ) e ^{ x }+ A
$
Comparing the coefficients of $e ^{ x }$ and constant term on both sides, we get
$
4 A+4 B=2 \text { and } A=-3
$
Solving these equations, we get
$
\begin{aligned}
& B=\frac{7}{2} \\
& \therefore I=\frac{-3\left(4 e^x+1\right)+\frac{7}{2}\left(4 e^x\right)}{4 e^x+1} d x \\
& =-3 \int d x+\frac{7}{2} \int \frac{4 e^x}{4 e^x+1} d x \\
& \therefore I=-3 x+\frac{7}{2} \log \left|4 e^x+1\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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