Question

\[\int\frac{\text{ log }\left( x + 2 \right)}{\left( x + 2 \right)^2}  \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\frac{\text{log }\left( x + 2 \right) dx}{\left( x + 2 \right)^2}\]
\[\text{ Let log }\left( x + 2 \right) = t\]
\[ \Rightarrow x + 2 = e^t \]
\[ \Rightarrow \frac{1}{\left( x + 2 \right)}dx = dt\]
\[ \therefore I = \int\frac{t}{e^t}dt\]
\[ = \int t e^{- t} dt\]
`  " Taking t as the first function and e"^- t" as the second function " . `
\[ = t\int e^{- t} - \int\left\{ \frac{d}{dt}\left( t \right)\int e^{- 2t} dt \right\}dt\]
\[ = t \times \frac{e^{- t}}{- 1} - \int1 \cdot e^{- t} dt\]
\[ =\text{  - t e}^{- t} + \frac{e^{- t}}{- 1} + C\]
\[ = - e^{- t} \left( t + 1 \right) + C\]
\[ = - \frac{\left( t + 1 \right)}{e^t} + C . . . (1)\]
\[\text{Substituting the value of t in eq} (1) \]
\[ = \frac{- \left[ \text{ log} \left( x + 2 \right) + 1 \right]}{x + 2} + C\]
\[ = - \frac{\text{ log } \left( x + 2 \right)}{x + 2} - \frac{1}{\left( x + 2 \right)} + C\]