If $\int\left(\frac{4 \mathrm{e}^x-25}{2 \mathrm{e}^x-5}\right) \mathrm{d} x=\mathrm{A}…

If $\int\left(\frac{4 \mathrm{e}^x-25}{2 \mathrm{e}^x-5}\right) \mathrm{d} x=\mathrm{A} x+\operatorname{Blog}\left(2 \mathrm{e}^x-5\right)+\mathrm{c} \quad$ (where c is a constant of integration) then
  1. $\mathrm{A}=5, \mathrm{~B}=3$
  2. $\mathrm{A}=5, \mathrm{~B}=-3$
  3. $\mathrm{A}=-5, \mathrm{~B}=3$
  4. $\mathrm{A}=-5, \mathrm{~B}=-3$

Solution

$\begin{aligned} \int \frac{4 \mathrm{e}^x-25}{2 \mathrm{e}^x-5} \mathrm{~d} x & =\int \frac{5\left(2 \mathrm{e}^x-5\right)-3\left(2 \mathrm{e}^x\right)}{2 \mathrm{e}^x-5} \mathrm{~d} x \\ & =5 \int \mathrm{~d} x-3 \int \frac{2 \mathrm{e}^x}{2 \mathrm{e}^x-5} \mathrm{~d} x \\ & =5 x-3 \log \left|2 \mathrm{e}^x-5\right|+\mathrm{c} \\ \therefore \quad \mathrm{A}=5 \text { and } \mathrm{B} & =-3\end{aligned}$

Asked in: MHT CET 2024 (04 May Shift 2)

Practice more Indefinite Integration questions on Aicharya