Question

$\int \frac{log(x+1)-logx}{x(x+1)}dx=$

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

Answer 1

The correct option is A $\frac{-1}{2}\left\{log\right(x+1{)}^2\}+\frac{-1}{2}(logx{)}^2+log(x+1).logx+C$

$I=\int \frac{log\left(\frac{x+1}{x}\right)}{x^2(1+\frac{1}{x})}dx\Rightarrow I=\int \frac{log(1+\frac{1}{x})}{x^2(1+\frac{1}{x})}\Rightarrow I=-\int log(1+\frac{1}{x})dlog(1+\frac{1}{x})\therefore I=-\frac{1}{2}{\left\{log(1+\frac{1}{x})\right\}}^2+c\Rightarrow I=-\frac{1}{2}\left\{log\right(x+1)-logx{\}}^2+c\therefore I=-\frac{1}{2}[\left\{log\right(x+1){\}}^2+(logx{)}^2-2log(x+1).logx]+c$