Question

Evaluate the integral $\int \frac{\log (x+1)-\log \left(x\right)}{x(x+1)}dx$

Answer 1

Solve the given integral:

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

Let $\log (x+1)-\log \left(x\right)=t$

Differentiate the above expression we get

$$\begin{gathered} \frac{d}{dx}\left[\log (x+1)-\log x\right]=\frac{d}{dx}t \\ \Rightarrow \left[\frac{1}{(x+1)}-\frac{1}{x}\right]=\frac{dt}{dx} \\ \Rightarrow \left[\frac{x-(x+1)}{x(x+1)}\right]dx=dt \\ \Rightarrow \left[\frac{-1}{x(x+1)}\right]dx=dt \end{gathered}$$

Now put the value of $\log (x+1)-\log x,\frac{-1}{x(x+1)}dx$ in the given expression

$$\begin{gathered} =\int -tdt \\ =\frac{-t^2}{2} \\ =-\frac{1}{2}{\left[\log (x+1)-\log \left(x\right)\right]}^2\left[\because t=\log (x+1)-\log \left(x\right)\right] \end{gathered}$$

Hence the value of the given integral is $-\frac{1}{2}{\left[\log (x+1)-\log \left(x\right)\right]}^2$