Misalkan $u=6x^2-4$ , maka $du=12x\ dx$ $\Leftrightarrow dx=\frac{du}{12x}$

Sehingga menjadi:

$\int\frac{\pi x}{\sqrt{6x^2-4}}dx=\pi\int\frac{1}{\sqrt{6x^2-4}}x\ dx$

$=\pi\int\frac{1}{\sqrt{u}}x\ \frac{du}{12x}$

$=\pi\int\frac{1}{\sqrt{u}}\frac{du}{12}$, ingat bahwa $\sqrt{x}=x^{\frac{1}{2\ }}$ dan $\frac{1}{x^n}=x^{-n}$

$=\frac{\pi}{12}\int u^{-\frac{1}{2}}du$, untuk $f\left(x\right)=ax^n,\ n\ne-1$ maka $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$

$=\frac{\pi}{12}\left(\frac{1}{-\frac{1}{2}+1}u^{-\frac{1}{2}+1}\right)+C$

$=\frac{\pi}{12}\left(\frac{1}{-\frac{1}{2}+\frac{2}{2}}u^{-\frac{1}{2}+\frac{2}{2}}\right)+C$

$=\frac{\pi}{12}\left(\frac{1}{\frac{1}{2}}u^{\frac{1}{2}}\right)+C$

$=\frac{\pi}{12}\left(2\sqrt{u}\right)+C$

$=\frac{\pi}{6}\sqrt{u}+C$

$=\frac{\pi}{6}\sqrt{6x^2-4}+C$

Jadi, hasil integral substitusi tersebut adalah $\frac{\pi}{6}\sqrt{6x^2-4}+C$