Question

\[\int\frac{x + \sqrt{x + 1}}{x + 2} dx\]

Answer 1

\[We\ have, \]
\[I = \int\frac{x + \sqrt{x + 1}}{x + 2} dx\]
\[\text{Let}, x + 1 = t^2 \]
\[\text{Differentiating both sides we get}\]
\[dx = 2tdt\]
\[\text{Now, integration becomes}\]
\[I = \int\frac{\left( t^2 - 1 + t \right)}{t^2 + 1}2t dt\]
\[ = 2\int\frac{t^3 + t^2 - t}{t^2 + 1} dt\]
\[ = 2\int\frac{t^3 + t - t + t^2 + 1 - 1 - t}{t^2 + 1} dt\]
\[ = 2\int\frac{t^3 + t + t^2 + 1 - t - t - 1}{t^2 + 1} dt\]
\[ = 2\int\frac{t^3 + t}{t^2 + 1} dt + + 2\int\frac{t^2 + 1}{t^2 + 1} dt + 2\int\frac{- 2t - 1}{t^2 + 1} dt\]
\[ = 2\  ∫  tdt + 2\  ∫ dt - 2\int\frac{2t}{t^2 + 1} dt - 2\int\frac{1}{t^2 + 1} dt\]
\[ = t^2 + \text{2t - 2}\text{log }\left| t^2 + 1 \right| - 2 \tan^{- 1} t + C\]
\[ = \left( x + 1 \right) + 2\sqrt{x + 1} - 2\text{log} \left| x + 2 \right| - 2 \tan^{- 1} \sqrt{x + 1} + C\]