Question

$\int e^{-x}\left[\frac{2x+5}{(2x+3{)}^2}\right]dx=$

• A. $\frac{e^{-x}}{2x+3}+c$
• B. $\frac{-e^{-x}}{2x+3}+c$
• C. $e^{-x}\frac{1}{2x+5}+c$
• D. $-e^{-x}\frac{1}{2x+5}+c$

Answer 1

The correct option is B $\frac{-e^{-x}}{2x+3}+c$

$\int e^{-x}[\frac{1}{(2x+3)}+\frac{2}{(2x+3{)}^2}]dx$

$-x=t$

$-\int \left[e^t\right(\frac{1}{-2t+3})+\frac{2}{(-2t+3{)}^2}]dt$

$+\int e^t\left[\right(\frac{1}{2t-3})-\frac{2}{(2t-3{)}^2}]dt$

In the form of

$\int e^t\left(f\right(t)+f^1(t\left)\right]dt=e^{t}f\left(t\right)+c$

$=\frac{e^t}{2t-3}+c=\frac{e^{-x}}{-2x-3}+c=\frac{-e^{-x}}{2x+3}+c$