Misalkan $u=5x^2-2$, maka $du=10x dx$ $\Leftrightarrow dx=\frac{du}{10x}$

Sehingga menjadi:

$\int x\sqrt{5x^2-2}dx=\int\left(5x^2-2\right)^{\frac{1}{2}}x\ dx$

$=\int u^{\frac{1}{2}}x\ \frac{du}{10x}$

$=\int u^{\frac{1}{2}}\frac{du}{10}$

$=\frac{1}{10}\int u^{\frac{1}{2}}du$, untuk $f\left(x\right)=ax^n,\ n\ne-1$ maka $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$

$=\frac{1}{10}\left(\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}\right)+C$

$=\frac{1}{10}\left(\frac{1}{\frac{1}{2}+\frac{2}{2}}u^{\frac{1}{2}+\frac{2}{2}}\right)+C$

$=\frac{1}{10}\left(\frac{1}{\frac{3}{2}}u^{\frac{3}{2}}\right)+C$

$=\frac{1}{10}\left(\frac{2}{3}u^{\frac{3}{2}}\right)+C$

$=\frac{1}{15}\left(u^{\frac{2}{2}}\right)\left(u^{\frac{1}{2}}\right)+C$

$=\frac{1}{15}u\sqrt{u}+C$

$=\frac{1}{15}\left(5x^2-2\right)\sqrt{5x^2-2}+C$

Jadi, hasil integral substitusi tersebut adalah $\frac{1}{15}\left(5x^2-2\right)\sqrt{5x^2-2}+C$