Question

Evaluate $\int e^x(\log \left(x\right)+\frac{1}{x^2})dx$

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

Answer 1

The correct option is D $e^x(\log x-\frac{1}{x})$

Given,

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

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

We know,

$\int u.vdx=u\int vdx$ $-\int (\frac{du}{dx}\int vdx)dx$

$=(e^x\log \left(x\right)-\text{Ei}\left(x\right))-\frac{e^x}{x}+\text{Ei}\left(x\right)$ where $\text{Ei}\left(x\right)=\int \frac{e^x}{x}dx$

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

$=e^x(\log x-\frac{1}{x})$