Nilai integral 36x^3 / akar 17-3x^4 dx adalah..

[tex]\displaystyle \int \frac{36x^3}{\sqrt{17-3x^4}}$ dx$ \\ $Misalkan :$ \\ u=17-3x^4 \\ du=-12x^3$ dx$ \\ $Sehingga,$ \\ 36x^3$ dx$=-3du \\ \displaystyle \int \frac{-3du}{\sqrt{u}}=\int \frac{-3}{\sqrt{u}}du \\ \int -3u^{-\frac{1}{2}}du=\frac{-3u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C=\frac{-3u^{\frac{1}{2}}}{\frac{1}{2}}+C=-6\sqrt{u}+C \\ \\ $Substitusi lagi, menjadi:$ \\ \displaystyle \int \frac{36x^3}{\sqrt{17-3x^4}}$ dx$ =-6\sqrt{17-3x^4}+C[/tex]