Question

$\int \frac{\log \left(x+2\right)}{{\left(x+2\right)}^2}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{\log \left(x+2\right)dx}{{\left(x+2\right)}^2} \\ \mathrm{Let}\log \left(x+2\right)=t \\ \Rightarrow x+2=e^t \\ \Rightarrow \frac{1}{\left(x+2\right)}dx=dt \\ \therefore I=\int \frac{t}{e^t}dt \\ =\int t{e}^{-t}dt \\ \mathrm{Taking}t\mathrm{as}\mathrm{the}\mathrm{first}\mathrm{function}\mathrm{and}{e}^{-t}\mathrm{as}\mathrm{the}\mathrm{second}\mathrm{function}. \\ =t\int e^{-t}-\int \left\{\frac{d}{dt}\left(t\right)\int e^{-2t}dt\right\}dt \\ =t\times \frac{e^{-t}}{-1}-\int 1·e^{-t}dt \\ =-t{e}^{-t}+\frac{e^{-t}}{-1}+C \\ =-e^{-t}\left(t+1\right)+C \\ =-\frac{\left(t+1\right)}{e^t}+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{.}\mathit{.}.\left(1\right) \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}t\mathit{}\mathrm{in}\mathrm{eq}\left(1\right) \\ =\frac{-\left[\log \left(x+2\right)+1\right]}{x+2}+C \\ =-\frac{\log \left(x+2\right)}{x+2}-\frac{1}{\left(x+2\right)}+C \end{gathered}$$