Question

$\int \frac{e^x(x-1)}{x^2}dx=$

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

Answer 1

The correct option is A $\frac{1}{x}{e}^x+c$

$\int \frac{e^x(x-1)}{x^2}dx?$

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

Let $f\left(x\right)=\frac{1}{x}$

$f^{\prime }\left(x\right)=\frac{-1}{x^2}$

we know that $\int e^x\left(f\right(x)+f^{\prime }(x\left)\right)dx=e^{x}f\left(x\right)+c$

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

$=\frac{1e^x}{x}+c$