Soal UN 2015 IPA
Tentukan \( \int 3x^2 \sqrt{(2x^3+5)} \ dx = \cdots \ ? \)
- \( \frac{3}{4}(2x^3+5) \sqrt{(2x^3+5)} + C \)
- \( \frac{1}{2}(2x^3+5) \sqrt{(2x^3+5)} + C \)
- \( \frac{2}{5}(2x^3+5) \sqrt{(2x^3+5)} + C \)
- \( \frac{1}{3}(2x^3+5) \sqrt{(2x^3+5)} + C \)
- \( \frac{1}{6}(2x^3+5) \sqrt{(2x^3+5)} + C \)
Pembahasan:
Misalkan \(u = 2x^3+5\) sehingga diperoleh:
\begin{aligned} u=2x^3+5 \Leftrightarrow \frac{du}{dx} &= 6x^2 \\[8pt] dx &= \frac{1}{6x^2} \ du \end{aligned}
Selanjutnya, substitusi hasil yang diperoleh di atas ke soal integral, diperoleh:
\begin{aligned} \int 3x^2 \sqrt{(2x^3+5)} \ dx &= \int 3x^2 \ \sqrt{u} \cdot \frac{1}{6x^2} \ du \\[8pt] &= \frac{3}{6} \int \sqrt{u} \ du = \frac{1}{2} \int u^{\frac{1}{2}} \ du \\[8pt] &= \frac{1}{2} \cdot \frac{2}{3}u^{\frac{3}{2}} + C \\[8pt] &= \frac{1}{3}(2x^3+5)^{\frac{3}{2}} + C \\[8pt] &= \frac{1}{3} (2x^3+5) \sqrt{(2x^3+5)} + C \end{aligned}
Jawaban D.