<s>[INST] How do I show equivalence between $\frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3}$ and $\frac{(x^2 + 1)^{3/2}(3x^2 - 2)}{15}$? I wanted to find $\int x^3 \sqrt{x^2 + 1} dx$ and came to this answer:
$$\frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3}$$
Using software to verify, I find Maple does not show equivalence between this and the actual answer, yet an online integral calculator does show equivalence between this and the actual answer with a footnote that further simplification does yield the expected answer.
For reference, the actual answer is:
$$\frac{(x^2 + 1)^{3/2}(3x^2 - 2)}{15}$$
How do I show algebraically that these two are equivalent? I could plug in $x$ for an arbitrary amount of numbers and compare but I want to be more declarative and show equality algebraically.
I've tried to combine this into one fraction:
$$\begin{align}
 & \frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3} \\
=& \frac{3\times(x^2 + 1)^{5/2}}{3\times5} - \frac{5\times(x^2 + 1)^{3/2}}{5\times3} \\
=& \frac{3(x^2 + 1)^{5/2}}{15} - \frac{5(x^2 + 1)^{3/2}}{15} \\
=& \frac{3(x^2+1)^{5/2} - 5(x^2+1)^{3/2}}{15}
\end{align}$$
But I am not sure where or how else to go from here.
 [/INST] Factor out $(x^2+1)^{3/2}$:
$$\frac{(x^2 + 1)^{5/2}}{5} - \frac{(x^2 + 1)^{3/2}}{3}$$
$$=(x^2+1)^{3/2}\left(\dfrac{x^2+1}5-\dfrac13\right)$$
$$=(x^2+1)^{3/2}\;\dfrac{[3(x^2+1)-5]}{15}$$
$$=(x^2+1)^{3/2}\;\dfrac{(3x^2-2)}{15}$$
Or from where you left off:
$$\frac{3(x^2+1)^{5/2} - 5(x^2+1)^{3/2}}{15}$$
$$=(x^2+1)^{3/2}\;\dfrac{[3(x^2+1)-5]}{15}$$
$$=(x^2+1)^{3/2}\;\dfrac{(3x^2-2)}{15}$$
</s>