Question

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

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

Answer 1

Answer: (B)

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

$\int e^x(logx+\frac{1}{x^2})dx$

$\int e^{x}logxdx+\int e^x\frac{1}{x^2}dx$

$=logx\int e^{x}dx-\int \frac{1}{x}{e}^{x}dx+\int \frac{e^x}{x^2}dx$

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

$=logx{e}^x-\int \frac{1}{x}{e}^{x}dx+e^x[-\frac{1}{x}]+\int e^x\frac{1}{x}dx$

$=e^x[logx-\frac{1}{x}]+c$

So, $\int e^x(logx+\frac{1}{x^2})dx=e^x[logx-\frac{1}{x}]+c$