Discussion Forum

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

Answer:

Explanation:

Let   l=$\int\left(\frac{4e^{x}-25}{2e^{x}-5}\right)dx$

 =  $\int\frac{4e^{x}}{2e^{x}-5}dx-\int\frac{25}{2e^{x}-5}dx$

 $=4\int\frac{e^{x}}{2e^{x}-5}dx-25\int\frac{e^{-x}}{2-5e^{-x}}dx$

 Put $2e^{x}-5=u$ and 2-5$e^{-x}$=v

$\Rightarrow$    $ 2e^{x} dx$=du    and  5$e^{-x}dx= du$

$\Rightarrow e^{x}dx=\frac{du}{2} $  and    $ e^{-x}dx=\frac{dv}{5}$

$\therefore$    $l=4\int_{}^{} \frac{du}{2u}-25\int\frac{du}{5v}$

  = 2 log u-5 log v+c

 $=2\log(2e^{x}-5)-5\log(2-5e^{x})+c$

$=2\log(2e^{x}-5)-5\log\left(\frac{2e^{x}-5}{e^{x}}\right)+c$

 $=2\log(2e^{x}-5)-5\log(2e^{x}-5)+5\log e^{x}+c$

$=-3\log(2e^{x}-5)+5x+c$

 $\Rightarrow l=5x-3\log(2e^{x}-5)+c$

 But it is given l=Ax+Blog (2e^x -5)+c

$\therefore$   A=5  and B=-3