Question

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

Answer 1

\[\text{ We  have,} \]
\[I = \int \frac{dx}{\left( x^2 + 1 \right) \sqrt{x}}\]
\[\text{ Putting  x }= t^2 \]
\[dx = 2t \text{ dt }\]
\[ \therefore I = \int \frac{2t \text{ dt }}{\left[ \left( t^2 \right)^2 + 1 \right]t}\]
\[ = 2\int \frac{dt}{t^4 + 1}\]
\[ = \int \left[ \frac{\left( t^2 + 1 \right) - \left( t^2 - 1 \right)}{\left( t^4 + 1 \right)} \right]dt\]
\[ = \int\left( \frac{t^2 + 1}{t^4 + 1} \right)dt - \int\left( \frac{t^2 - 1}{t^4 + 1} \right)dt\]
\[\text{Dividing numerator & denominator by }t^2 \]
\[I = \int\left( \frac{1 + \frac{1}{t^2}}{t^2 + \frac{1}{t^2}} \right)dt - \int \frac{\left( 1 - \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2}}\]
\[ = \int \frac{\left( 1 + \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} - 2 + 2} - \int \frac{\left( 1 - \frac{1}{t^2} \right)dt}{t^2 + \frac{1}{t^2} + 2 - 2}\]
\[ = \int \frac{\left( 1 + \frac{1}{t^2} \right)dt}{\left( t - \frac{1}{t} \right)^2 + \left( \sqrt{2} \right)^2} - \int \frac{\left( 1 - \frac{1}{t^2} \right)dt}{\left( t + \frac{1}{t} \right)^2 - \left( \sqrt{2} \right)^2}\]
\[\text{ Putting t }- \frac{1}{t} = p\]
\[ \Rightarrow \left( 1 + \frac{1}{t^2} \right)dt = dp\]
\[\text{ Putting  t }+ \frac{1}{t} = q\]
\[ \Rightarrow \left( 1 - \frac{1}{t^2} \right)dt = dq\]
\[ \therefore I = \int\frac{dp}{p^2 + \left( \sqrt{2} \right)^2} - \int\frac{dq}{q^2 - \left( \sqrt{2} \right)^2}\]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{p}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\text{ log }\left| \frac{q - \sqrt{2}}{q + \sqrt{2}} \right| + C\]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{t - \frac{1}{t}}{\sqrt{2}} \right) - \frac{1}{2\sqrt{2}}\text{ log }\left| \frac{t + \frac{1}{t} - \sqrt{2}}{t + \frac{1}{t} + \sqrt{2}} \right| + C\]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{t^2 - 1}{\sqrt{2}t} \right) - \frac{1}{2\sqrt{2}}\text{ log }\left| \frac{t^2 - \sqrt{2}t + 1}{t^2 + \sqrt{2}t + 1} \right| + C\]
\[ = \frac{1}{\sqrt{2}} \tan^{- 1} \left( \frac{x - 1}{\sqrt{2x}} \right) - \frac{1}{2\sqrt{2}}\text{ log} \left| \frac{x - \sqrt{2x} + 1}{x + \sqrt{2x} + 1} \right| + C\]