Question

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