Question

The value of the integral $\int \frac{\log (x+1)-\log x}{x(x+1)}dx$ is

• A. $-\frac{1}{2}\left[\log \right(x+1){]}^2-\frac{1}{2}(\log x{)}^2+\log (x+1)\log x+C$
• B. $-\left[\log \right(x+1{)}^2-\left(\log x{)}^2\right]+\log (x+1)\log x+C$
• C. $C-\left(1/2\right)\left[\log \right(1+1/x){]}^2$
• D. none of these

Answer 1

Answer: (A), (C)

The correct options are
A $-\frac{1}{2}\left[\log \right(x+1){]}^2-\frac{1}{2}(\log x{)}^2+\log (x+1)\log x+C$
C $C-\left(1/2\right)\left[\log \right(1+1/x){]}^2$

Let $I=\int \frac{\log (x+1)-\log x}{x(x+1)}dx$
Substitute $t=1+\frac{1}{x}\Rightarrow dt=-\frac{dx}{x^2}$
$I=-\int \frac{\log t}{t}dt=-\frac{1}{2}{\left(\log t\right)}^2+c$
$=-\frac{1}{2}\left({\left(\log 1+\frac{1}{x}\right)}^2\right)+c$
$=-\frac{1}{2}\left({\left(\log x+1\right)}^2\right)-\frac{1}{2}\left({\left(\log x\right)}^2\right)+\left(\log x+1\right)\log x+c$