Evalute : 3 e^ 2 t +5 4 e^ 2 t -5 d t - Vidyadip

Question

Evalute : $\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t$

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Answer

Let $I =\int \frac{3 e^{2 t}+5}{4 e^{2 t}-5} d t$
Put, Numerator $= A ($ Denominator $)+ B \left[\frac{d}{d x}\right.$ (Denominator $\left.)\right]$
$
\begin{aligned}
& \therefore 3 e ^{2 t }+5= A \left(4 e ^{2 t }-5\right)+ B \left[\frac{d}{d t}\left(4 e ^{2 t }-5\right)\right] \\
& \therefore 3 e ^{2 t }+5= A \left(4 e ^{2 t }-5\right)+ B \left[4 e ^{2 t } \times 2-0\right] \\
& \therefore 3 e ^{2 t }+5=(4 A +8 B ) e ^{2 t }-5 A
\end{aligned}
$
Equating the coefficient of $e ^{2 t }$ and constant on both sides, we get
$
\begin{aligned}
& 4 A +8 B =3 \\
& \text { and }-5 A =5 \\
& \therefore A =-1 \\
& \therefore \text { from }(1), 4(-1)+8 B =3 \\
& \therefore 8 B =7 \\
& \therefore B =\frac{7}{8} \\
& \therefore 3 e^{2 t}+5=-\left(4 e^{2 t}-5\right)+\frac{7}{8}\left(8 e^{2 t}\right) \\
& \left.\therefore I=\int \frac{-\left(4 e^{2 t}-5\right)+\frac{7}{8}\left(8 e^{2 t}\right)}{4 e^{2 t}-5}\right] d t \\
& \quad=\int\left[-1+\frac{\frac{7}{8}\left(8 e^{2 t}\right)}{4 e^{2 t}-5}\right] d t \\
& \quad=-\int 1 d t+\frac{7}{8} \int \frac{8 e^{2 t}}{4 e^{2 t}-5} d t \\
& \quad=-t+\frac{7}{8} \log \left|4 e^{2 t}-5\right|+c \cdot \\
& ......\left[\because \int \frac{f^{\prime}(x)}{f(x)} d x=\log |f(x)|+c\right] .
\end{aligned}
$

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