<s>[INST] Evaluation of $\int x^8\sqrt{x^2+1}\ dx$ 
Evaluation of $\displaystyle \int x^8\sqrt{x^2+1}\ dx$

$\bf{My\: Try::}$ Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d \theta$
So $$I = \int \sec ^{3}\theta \cdot \tan^{8}\theta d \theta=\int \frac{\sin^{8}\theta}{\cos^{5}\theta }d\theta$$
Now How can I solve it after that, Help Required, Thanks
 [/INST] Alternative Euler approach:
$$\begin{align*}
I &= \int x^8 \sqrt{x^2+1} \, dx \\
&= \frac19 x^9 \sqrt{x^2+1} - \frac19 \int \frac{x^{10}}{\sqrt{x^2+1}} \, dx \tag1\\
&=\frac19 x^9 \sqrt{x^2+1} - \frac1{18} \int \frac{y^5}{\sqrt{y^2+y}} \, dy \tag2\\
&=\frac19 x^9 \sqrt{x^2+1} - \frac19 \int \frac{z^9}{(1-2z)^6} \,dz \tag3\\
&=\frac19 x^9 \sqrt{x^2+1} + \frac1{9216} \int \frac{(1-w)^9}{w^6} \,dw \tag4\\
\end{align*}$$
and proceed with partial fraction expansion.


*

*$(1)$ : integrate by parts

*$(2)$ : substitute $y=x^2$

*$(3)$ : substitute $y=\dfrac{z^2}{1-2z}$

*$(4)$ : substitute $w=1-2z$
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