Question

$\int \frac{\log (x+1)-\log x}{x(x+1)}dx$ is equal to

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

Answer 1

The correct option is A $-\frac{1}{2}{\left[\log \left(\frac{x+1}{x}\right)\right]}^2+c$

We have,

$I=\int \frac{\log \left(x+1\right)-\log x}{x\left(x+1\right)}dx$

$I=\int \frac{\frac{\log \left(x+1\right)}{\log x}}{x\left(x+1\right)}dx$

Let

$\log \frac{\left(x+1\right)}{x}=p$

$\frac{x}{\left(x+1\right)}\frac{-1}{x^2}dx=dp$

$-\frac{dx}{x\left(x+1\right)}=dp$

Therefore,

$I=-\int pdp=-\frac{p^2}{2}+C$

$I=-\frac{1}{2}{\left[\log \left(\frac{x+1}{x}\right)\right]}^2+C$

Hence, this is the answer.