Question

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

Answer 1

\[\text{ We  have,} \]
\[I = \int \frac{dx}{\left( x + 1 \right) \sqrt{x^2 + x + 1}}\]
\[\text{ Putting  x }+ 1 = \frac{1}{t}\]
\[ \Rightarrow dx = - \frac{1}{t^2}dt\]
\[ \therefore I = \int \frac{- \frac{1}{t^2}dt}{\frac{1}{t}\sqrt{\left( \frac{1}{t}, - , 1 \right)^2 + - 1 + 1\frac{1}{t}}}\]
\[ = \int \frac{- \frac{1}{t^2}dt}{\frac{1}{t}\sqrt{\frac{1}{t^2} - + 1 + \frac{2}{t}\frac{1}{t}}}\]
\[ = \int \frac{- \frac{1}{t}dt}{\frac{\sqrt{t^2 + t - 2t + 1}}{t}}\]
\[ = - \int \frac{dt}{\sqrt{t^2 - t + 1}}\]
\[ = - \int\frac{dt}{\sqrt{t^2 - t + \frac{1}{4} - \frac{1}{4} + 1}}\]
\[ = - \int\frac{dt}{\sqrt{\left( t - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2}}\]
\[ = - \text{ log }\left| t - \frac{1}{2} + \sqrt{\left( t - \frac{1}{2} \right)^2 + \left( \frac{\sqrt{3}}{2} \right)^2} \right| + C\]
\[ = - \text{ log }\left| t - \frac{1}{2} + \sqrt{t^2 - t + 1} \right| + C\]
\[ = - \text{ log }\left| \frac{1}{x + 1} - \frac{1}{2} + \sqrt{\frac{1}{\left( x + 1 \right)^2} - \frac{1}{x + 1} + 1} \right| + C\]
\[ = - \text{ log }\left| \frac{1}{x + 1} - \frac{1}{2} + \frac{\sqrt{\left( x + 1 \right)^2 - \left( x + 1 \right) + 1}}{x + 1} \right| + C\]
\[ = - \text{ log }\left| \frac{1}{x + 1} - \frac{1}{2} + \frac{\sqrt{x^2 + x + 1}}{x + 1} \right| + C\]