Question

$\int\left(\frac{4e^{2}-25}{2e^{x}-5}\right)dx = Ax+B \,\,log |2e^{x}-5|+c$ then

• A. $ A = 5$, and $B = 3$
• B. $ A = 5$, and $B = - 3 $
• C. $A = - 5$, and $B = 3$
• D. $ A = - 5$, and $B = - 3$

Answer 1

Answer: (B)

Let $I=\int\left(\frac{4 e^{x}-25}{2 e^{x}-5}\right) dx$
$=\int \frac{4 e^{x}}{2 e^{x}-5} d x-\int \frac{25}{2 e^{x}-5} d x $
$=4 \int \frac{e^{x}}{2 e^{x}-5} d x-25 \int \frac{e^{-x}}{2-5 e^{-x}} d x$
Put $2 e^{x}-5=u$ and $2-5 e^{-x}=v$
$\Rightarrow 2 e^{x} dx=du$
and $5 e^{-x} d x=dv$
$\Rightarrow e^{x} d x=\frac{du}{2}$ and $e^{-x} dx=\frac{dv}{5}$
$\therefore I=4 \int \frac{du}{2u}-25 \int \frac{du}{5v}$
$=2 \log u-5 \log v+c$
$=2 \log \left(2 e^{x}-5\right)-5 \log \left(2-5 e^{-x}\right)+c$
$=2 \log \left(2 e^{x}-5\right)-5 \log \left(\frac{2 e^{x}-5}{e^{x}}\right)+c$
$=2 \log \left(2 e^{x}-5\right)-5 \log \left(2 e^{x}-5\right)+5 \log e^{x}+c$
$=-3 \log \left(2 e^{x}-5\right)+5x+c$
$\Rightarrow I=5 x-3 \log \left(2 e^{x}-5\right)+c$
But it is given $I=A x+B \log \left(2 e^{x}-5\right)+c$
$\therefore A=5$ and $B=-3$