\textbf{Partial integration}

This basically allows you to swap what you integrate and what you integrate by.

,,\qquad\int u\ \mathrm{d}v = uv - \int v\ \mathrm{d}u + C


Usually, you use it to solve an integral with polynomial multiplied by one of the non-polynomial functions.

,,\qquad \int P_n(x)\cdot \begin{matrix}e^{ax}\\\sin(ax)\\\cos(ax)\\\ln(ax)\\\arcsin(ax)\\\arccos(ax)\end{matrix}\mathrm{d}x

You usually choose \[u\] to be the non-polynomial because calculating the derivative of it is probably going to be easier than integrating it.
On the other hand, polynomials are easy to both derive and integrate.

\textbf{Example}

You integrate  \[\int x^2\arccos(x)\mathrm{d}x\] by differentiating \[\arccos(x)\] and by integrating \[x^2\]:

,align \qquad \int \arccos(x)\cdot x^2 \mathrm{d}x &= \begin{Bmatrix} u = \arccos(x) & \mathrm{d}u = -\frac{\mathrm{d}x}{\sqrt{1- x^2}}\\\mathrm{d}v = x^2\mathrm{d}x & v = \frac{x^3}{3}\end{Bmatrix} = \frac{x^3}{3}\arccos(x)+ \frac{1}{3}\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x +C \\&= \frac{x^3}{3}\arccos(x) - \frac{1}{3}\left(\sqrt{1-x^2} + \frac{\sqrt{(1-x^2)^3}}{3}\right) + C

As for how you solve \[\int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x\] you do it by substituting \[t = \sqrt{1-x^2},\quad x^2 = 1- t^2,\quad \cancel{2}\mathrm{d}x = -\cancel{2}t\mathrm{d}t\]:

,,\qquad \int\frac{x^3}{\sqrt{1-x^2}}\mathrm{d}x = \int\frac{1-t^2}{t}(-t)\mathrm{d}t = t - \frac{t^3}{3} = \boxed{\sqrt{1-x^2} - \frac{\sqrt{(1-x^2)^3}}{3}}