Question

\[\int\frac{1}{\left( x + 1 \right)\left( x^2 + 2x + 2 \right)} dx\]

Answer 1

\[\text{Let I} = \int\frac{dx}{\left( x + 1 \right) \left( x^2 + 2x + 2 \right)}\]
\[ = \int\frac{dx}{\left( x + 1 \right) \left[ x^2 + 2x + 1 + 1 \right]}\]
\[ = \int\frac{dx}{\left( x + 1 \right) \left[ \left( x + 1 \right)^2 + 1 \right]}\]
\[\text{Putting}\ x + 1 = t\]
\[ \Rightarrow dx = dt\]
\[\text{Now, integral becomes}\]
\[I = \int\frac{dt}{t \left[ t^2 + 1 \right]}\]
\[ = \int\frac{t \cdot dt}{t^2 \left( t^2 + 1 \right)}\]
\[\text{Again putting }t^2 = p\]
\[ \Rightarrow \text{2t dt }= dp\]
\[ \Rightarrow t dt = \frac{dp}{2}\]
\[\text{Now, integral becomes}\]
\[I = \frac{1}{2} \int\frac{dp}{p \left( p + 1 \right)}\]
\[ = \frac{1}{2}\int\frac{dp}{p^2 + p}\]
\[ = \frac{1}{2}\int\frac{dp}{p^2 + p + \frac{1}{4} - \frac{1}{4}}\]
\[ = \frac{1}{2}\int\frac{dp}{\left( p + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}\]
\[ = \frac{1}{2} \left[ \frac{1}{2 \times \frac{1}{2}} \text{log }\left| \frac{p + \frac{1}{2} - \frac{1}{2}}{p + \frac{1}{2} + \frac{1}{2}} \right| \right] + C\]
\[ = \frac{1}{2} \text{log }\left| \frac{p}{p + 1} \right| + C\]
\[ = \frac{1}{2} \text{log }\left| \frac{t^2}{t^2 + 1} \right| + C\]
\[ = \frac{1}{2} \text{log }\left| \frac{\left( x + 1 \right)^2}{\left( x + 1 \right)^2 + 1} \right| + C\]
\[ = \text{log}\sqrt{\left| \frac{\left( x + 1 \right)^2}{\left( x + 1 \right)^2 + 1} \right|} + C\]
\[ = \text{log }\left| \frac{x + 1}{\sqrt{x^2 + 2x + 2}} \right| + C\]