Question

\[\int x\sqrt{x^4 + 1} \text{ dx}\]

Answer 1

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