Solve Q 6(iii)

Dear student,

Solving without limits,

$$\begin{gathered} \mathrm{Let} \mathrm{I} = \int \frac{\log \mathrm{x}}{{\left(\mathrm{x}+1\right)}^2} \mathrm{dx} \\ =\log \mathrm{x} \times \int \frac{\mathrm{dx}}{{\left(\mathrm{x}+1\right)}^2} - \int \left[\frac{\mathrm{d}}{\mathrm{dx}}\log \mathrm{x} \times \int \frac{\mathrm{dx}}{{\left(\mathrm{x}+1\right)}^2} \right]\mathrm{dx} \\ =-\frac{\log \mathrm{x}}{\left(\mathrm{x}+1\right)} + \int \frac{\mathrm{dx}}{\mathrm{x}\left(\mathrm{x}+1\right)} \\ =-\frac{\log \mathrm{x}}{\left(\mathrm{x}+1\right)} + \int \frac{\mathrm{dx}}{{\mathrm{x}}^2+\mathrm{x}+{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{4}}} \\ =-\frac{\log \mathrm{x}}{\left(\mathrm{x}+1\right)} + \int \frac{\mathrm{dx}}{{\left(\mathrm{x}+{\displaystyle \frac{1}{2}}\right)}^2 - {\left({\displaystyle \frac{1}{2}}\right)}^2} \\ =-\frac{\log \mathrm{x}}{\left(\mathrm{x}+1\right)} + \log \left|\frac{\mathrm{x}}{\mathrm{x}+1}\right| + \mathrm{C} \end{gathered}$$

Applying limits,