Question

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

• A. $\frac{1}{logx+2}+c$
• B. $\frac{x}{\log x+2}+c$
• C. $\frac{x}{\log x+1}+c$
• D. $\frac{1}{\log x+3}+c$

Answer 1

The correct option is B $\frac{x}{\log x+2}+c$

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

$\int \left[\frac{(logx+2)-1}{(logx+2{)}^2}\right]dx$

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

put $t=logx$ then

It is in the form of

$\int e^t\left(f\right(t)+f^{\prime }(t\left)\right)=e^{t}f\left(t\right)+c$

Where $f\left(t\right)=\frac{1}{t+2}$ Simplifing we get the answer

$=\frac{x}{(logx+2)}+c$