$\int e^x\left(\frac{x-1}{x^2}\right) d x=$

$\int e^x\left(\frac{x-1}{x^2}\right) d x=$
  1. $\frac{-e^x}{x^2}+c$
  2. $\frac{-e^x}{x}+c$
  3. $\frac{e^x}{x^2}+c$
  4. $\frac{e^x}{x}+c$

Solution

Let $I=\int e^x\left(\frac{x-1}{x^2}\right) d x=\int e^x\left(\frac{1}{x}-\frac{1}{x^2}\right) d x=e^x\left(\frac{1}{x}\right)+c$

Asked in: MHT CET 2021 (22 Sep Shift 1)

Practice more Indefinite Integration questions on Aicharya