How do you integrate the following integral?
$\int{x\sqrt{{{x}^{4}}+{{a}^{4}}}dx}$

Hint: This integration requires to be solved by two methods, the substitution and the by parts method. At first, we substitute ${{x}^{2}}$ with ${{a}^{2}}\tan \theta $ . After solving, we end up with the integral \[\dfrac{{{a}^{4}}}{2}\int{{{\sec }^{3}}\theta d\theta }\] . This has to be solved using the by parts method. We take $u=\sec \theta ,v={{\sec }^{2}}\theta $ and then finally solve it to get the answer.

Complete step-by-step solution:
The given integral that we have in this problem is,
$\int{x\sqrt{{{x}^{4}}+{{a}^{4}}}dx}=I$ (say)
We will solve it by substitution method. In this method, we substitute some expression of the variable of the integral by another variable in order to reduce the integral to a simpler form in order to solve it. Let us take,
$\begin{align}
  & {{x}^{2}}={{a}^{2}}\tan \theta ....\left( 1 \right) \\
 & \Rightarrow 2xdx={{a}^{2}}{{\sec }^{2}}\theta d\theta \\
 & \Rightarrow xdx=\dfrac{{{a}^{2}}}{2}{{\sec }^{2}}\theta d\theta \\
\end{align}$
Then, the integral becomes,
$\Rightarrow I=\dfrac{{{a}^{2}}}{2}\int{\left( \sqrt{{{\left( {{a}^{2}}\tan \theta \right)}^{2}}+{{a}^{4}}} \right){{\sec }^{2}}\theta d\theta }$
Upon opening up the brackets and simplifying the integral, we get,
 $\Rightarrow I=\dfrac{{{a}^{2}}}{2}\int{\left( \sqrt{{{a}^{4}}{{\tan }^{2}}\theta +{{a}^{4}}} \right){{\sec }^{2}}\theta d\theta }$
Taking ${{a}^{4}}$ outside of the square root, we get,
 \[\Rightarrow I=\dfrac{{{a}^{4}}}{2}\int{\left( \sqrt{{{\tan }^{2}}\theta +1} \right){{\sec }^{2}}\theta d\theta }\]
Using the formula ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ in the above integral, we get
\[\begin{align}
  & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\int{\left( \sqrt{{{\sec }^{2}}\theta } \right){{\sec }^{2}}\theta d\theta } \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\int{{{\sec }^{3}}\theta d\theta }....\left( 2 \right) \\
\end{align}\]
Now, we have to evaluate the above integral using the integration by parts method. In this method, we consider the integrand as product of two variable u and v like, $I=\int{uv}$ and the formula to solve this integral is,
$I=u\int{v}-\int{\left( \dfrac{du}{dx}\int{v} \right)}$
So, in the integral $\left( 2 \right)$ , taking $u=\sec \theta ,v={{\sec }^{2}}\theta $ , we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\int{\sec \theta \times {{\sec }^{2}}\theta d\theta } \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \int{{{\sec }^{2}}\theta d\theta }-\int{\left( \dfrac{d\left( \sec \theta \right)}{dx}\int{{{\sec }^{2}}\theta d\theta } \right)} \right] \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta -\int{\left( \sec \theta \tan \theta \left( \tan \theta \right) \right)} \right] \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta -\int{\left( \sec \theta {{\tan }^{2}}\theta \right)d\theta } \right] \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta -\int{\sec \theta \left( {{\sec }^{2}}\theta -1 \right)d\theta } \right] \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta -\int{{{\sec }^{3}}\theta d\theta }+\int{\sec \theta d\theta } \right] \\
\end{align}\]
Now, the integral \[\int{{{\sec }^{3}}\theta d\theta }\] on the right-hand side of the above equation is exactly the integral $I$ . So, the equation becomes,
\[\begin{align}
  & \Rightarrow I=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta -I+\int{\sec \theta d\theta } \right] \\
 & \Rightarrow I\left( 1+\dfrac{{{a}^{4}}}{2} \right)=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta +\int{\sec \theta d\theta } \right] \\
 & \Rightarrow I\left( 1+\dfrac{{{a}^{4}}}{2} \right)=\dfrac{{{a}^{4}}}{2}\left[ \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right]+c \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{{{a}^{4}}+2}\left[ \sec \theta \tan \theta +\ln \left| \sec \theta +\tan \theta \right| \right]+c....\left( 3 \right) \\
\end{align}\]
 From equation $\left( 1 \right)$ , we get,
$\tan \theta ={{\left( \dfrac{x}{a} \right)}^{2}},\sec \theta =\sqrt{1+{{\left( \dfrac{x}{a} \right)}^{2}}}=\dfrac{\sqrt{{{a}^{2}}+{{x}^{2}}}}{a}$
Replacing these in the equation $\left( 3 \right)$ , we get,
\[\begin{align}
  & \Rightarrow I=\dfrac{{{a}^{4}}}{{{a}^{4}}+2}\left[ \dfrac{\sqrt{{{a}^{2}}+{{x}^{2}}}}{a}\times {{\left( \dfrac{x}{a} \right)}^{2}}+\ln \left| \dfrac{\sqrt{{{a}^{2}}+{{x}^{2}}}}{a}+{{\left( \dfrac{x}{a} \right)}^{2}} \right| \right]+c \\
 & \Rightarrow I=\dfrac{{{a}^{4}}}{{{a}^{4}}+2}\left[ \dfrac{{{x}^{2}}\sqrt{{{a}^{2}}+{{x}^{2}}}}{{{a}^{3}}}+\ln \left| \dfrac{\sqrt{{{a}^{2}}+{{x}^{2}}}}{a}+\dfrac{{{x}^{2}}}{{{a}^{2}}} \right| \right]+c \\
\end{align}\]
Therefore, we can conclude that the integral in the problem can be solved to \[\dfrac{{{a}^{4}}}{{{a}^{4}}+2}\left[ \dfrac{{{x}^{2}}\sqrt{{{a}^{2}}+{{x}^{2}}}}{{{a}^{3}}}+\ln \left| \dfrac{\sqrt{{{a}^{2}}+{{x}^{2}}}}{a}+\dfrac{{{x}^{2}}}{{{a}^{2}}} \right| \right]+c\] .

Note: This integration seems to be very lengthy and it is. Also, this method includes two different methods, the substitution and the by parts methods. So, this requires separate dedication and effort. We should always be careful with the final conversion of the substitute variables with the substituted variables.