Question

$\int [\frac{1}{\log x}-\frac{1}{(\log x{)}^2}]dx=$

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

Answer 1

The correct option is B $\frac{x}{\log x}+c$

The integral $\int [\frac{1}{logx}-\frac{1}{(logx{)}^2}]dx$

is in the form of

$\int \left[x{f}^{\prime }\right(x)+f(x\left)\right]dx=xf\left(x\right)+c$

Hence the value is

$=\frac{x}{logx}+c$