3 e ^ 2 t +5 4 e ^ 2 t -5 dt

Question

$\int \frac{3 e ^{2 t }+5}{4 e ^{2 t }-5} dt$

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Answer

Let $I =\int \frac{3 e ^{2 t }+5}{4 e ^{2 t }-5} dt$
Let $3 e ^{2 t }+5= A \left(4 e ^{2 t }-5\right)+ B \frac{ d }{ dt }\left(4 e ^{2 t }-5\right)$
$=4 A e^{2 t}-5 A+B\left(8 e^{2 t}\right)$
$\therefore 3 e ^{2 t }+5=(4 A +8 B ) e ^{2 t }-5 A$
Comparing the coefficients of $e ^{2 t }$ and constant term on both sides,
we get $4 A+8 B=3$ and $-5 A=5$
Solving these equations,
we get $A=-1$ and $B=\frac{7}{8}$
$\therefore I=\int \frac{-1\left(4 e ^{2 t }-5\right)+\frac{7}{8}\left(8 e ^{2 t }\right)}{4 e ^{2 t }-5} dt$
$= \int dt +\frac{7}{8} \int \frac{8 e ^{2 t }}{4 e ^{2 t }-5} dt$
$\therefore I =- t +\frac{7}{8} \log \left|4 e ^{2 t }-5\right|+ c \quad \ldots \ldots . .\left[\frac{ f ^{\prime}(x)}{ f (x)} d x=\log | f (x)|+ c \right]$

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