Soal UN MTK IPA 2016
Hasil dari \( \int \frac{x^2-2}{\sqrt{6x-x^3}} \ dx = \cdots \)
- \( -\frac{3}{2} \sqrt{6x-x^3} + C \)
- \( -\frac{2}{3} \sqrt{6x-x^3} + C \)
- \( -\frac{1}{6} \sqrt{6x-x^3} + C \)
- \( \frac{1}{6} \sqrt{6x-x^3} + C \)
- \( \frac{2}{3} \sqrt{6x-x^3} + C \)
Pembahasan:
Misalkan \( u = 6x-x^3 \) sehingga diperoleh:
\begin{aligned} u = 6x-x^3 &\Leftrightarrow \frac{du}{dx} = 6-3x^2 \\[8pt] &\Leftrightarrow \frac{du}{dx} = -3(x^2-2) \\[8pt] &\Leftrightarrow dx = \frac{1}{-3(x^2-2)} \ du \end{aligned}
Selanjutnya, substitusi hasil di atas ke soal integral, diperoleh:
\begin{aligned} \int \frac{x^2-2}{\sqrt{6x-x^3}} \ dx &= \int \frac{x^2-2}{\sqrt{u}} \cdot \frac{1}{-3(x^2-2)} \ du \\[8pt] &= -\frac{1}{3} \int \frac{1}{\sqrt{u}} \ du = -\frac{1}{3} \int u^{-\frac{1}{2}} \ du \\[8pt] &= -\frac{1}{3} \cdot \frac{1}{-\frac{1}{2}+1}u^{-\frac{1}{2}+1} + C \\[8pt] &= -\frac{1}{3} \cdot 2u^\frac{1}{2} + C = -\frac{2}{3}\sqrt{u}+C \\[8pt] &= -\frac{2}{3} \sqrt{6x-x^3}+C \end{aligned}
Jawaban B.