Misalkan $u=t^4+2\pi t^2+12t$, maka $du=4t^3+4\pi t+12\ dt$ $\Leftrightarrow dt=\frac{du}{4t^3+4\pi t+12}$

Sehingga menjadi:

$\int\frac{t^3+\pi t+3}{\sqrt{t^4+2\pi t^2+12t}}dt=\int\frac{1}{\sqrt{t^4+2\pi t^2+12t}}\left(t^3+\pi t+3\right)dt$

$=\int\frac{1}{\sqrt{u}}\left(t^3+\pi t+3\right)\frac{du}{4t^3+4\pi t+12}$

$=\int\frac{1}{\sqrt{u}}\left(t^3+\pi t+3\right)\frac{du}{4\left(t^3+\pi t+3\right)}$

$=\int\frac{1}{\sqrt{u}}\frac{du}{4}$, ingat bahwa $\sqrt{x}=x^{\frac{1}{2}}$ dan $\frac{1}{x^n}=x^{-n}$

$=\frac{1}{4}\int u^{\ -\frac{1}{2}}du$, untuk $f\left(x\right)=ax^n,\ n\ne-1$ maka $\int ax^ndx=\frac{a}{n+1}x^{n+1}+C$

$=\frac{1}{4}\left(\frac{1}{-\frac{1}{2}+1}u^{\ -\frac{1}{2}+1}\right)+C$

$=\frac{1}{4}\left(\frac{1}{-\frac{1}{2}+\frac{2}{2}}u^{\ -\frac{1}{2}+\frac{2}{2}}\right)+C$

$=\frac{1}{4}\left(\frac{1}{\frac{1}{2}}u^{\ \frac{1}{2}}\right)+C$

$$$=\frac{1}{4}\left(2u^{\ \frac{1}{2}}\right)+C$

$=\frac{2}{4}\left(u^{\ \frac{1}{2}}\right)+C$

$=\frac{1}{2}\sqrt{u}+C$

$=\frac{1}{2}\sqrt{t^4+2\pi t^2+12t}+C$

Jadi, hasil integral substitusi tersebut adalah $\frac{1}{2}\sqrt{t^4+2\pi t^2+12t}+C$