Misalkan $u=2x^6-3x$, maka $du=12x^5-3 dx$ $\Leftrightarrow dx=\frac{du}{12x^5-3}$

Sehingga menjadi:

$\int\left(4x^5-1\right)\sqrt{2x^6-3x}\ dx=\int\left(2x^6-3x\right)^{\frac{1}{2}}\ \left(4x^5-1\right)dx$

$=\int u^{\frac{1}{2}}\ \left(4x^5-1\right)\frac{du}{12x^5-3}$

$=\int u^{\frac{1}{2}}\ \left(4x^5-1\right)\frac{du}{3\left(4x^5-1\right)}$

$=\int u^{\frac{1}{2}}\ \frac{du}{3}$

$=\frac{1}{3}\int u^{\frac{1}{2}}\ du$, untuk $f\left(x\right)=ax^n,\ n\ne-1$ maka $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$

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

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

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

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

$=\frac{2}{9}\left(u^{\frac{2}{2}}\right)\left(u^{\frac{1}{2}}\right)+C$

$=\frac{2}{9}u\sqrt{u}+C$

$=\frac{2}{9}\left(2x^6-3x\right)\sqrt{2x^6-3x}+C$

Jadi, hasil integral substitusi tersebut adalah $\frac{2}{9}\left(2x^6-3x\right)\sqrt{2x^6-3x}+C$