Soal UN MTK IPA 2018
Hasil dari \( \int \frac{x-1}{\sqrt{x^2-2x+10}} \ dx = \cdots \)
- \( -\sqrt{x^2-2x+10} + C \)
- \( -\frac{1}{2}\sqrt{x^2-2x+10} + C \)
- \( \frac{1}{2} \sqrt{x^2-2x+10} + C \)
- \( \sqrt{x^2-2x+10} + C \)
- \( 2\sqrt{x^2-2x+10} + C \)
Pembahasan:
Misalkan \( u = x^2-2x+10 \) sehingga diperoleh:
\begin{aligned} u = x^2-2x+10 &\Leftrightarrow \frac{du}{dx} = 2x-2 \\[8pt] &\Leftrightarrow dx = \frac{1}{2(x-1)} \ du \end{aligned}
Selanjutnya, substitusi hasil di atas ke soal integral, diperoleh:
\begin{aligned} \int \frac{x-1}{\sqrt{x^2-2x+10}} \ dx &= \int \frac{x-1}{\sqrt{u}} \cdot \frac{1}{2(x-1)} \ du \\[8pt] &= \frac{1}{2} \int \frac{1}{\sqrt{u}} \ du = \frac{1}{2} \int u^{-\frac{1}{2}} \ du \\[8pt] &= \frac{1}{2} \cdot \frac{1}{-\frac{1}{2}+1}u^{-\frac{1}{2}+1} + C \\[8pt] &= \frac{1}{2}\cdot 2u^{\frac{1}{2}} + C = \sqrt{u}+C \\[8pt] &= \sqrt{x^2-2x+10} + C \end{aligned}
Jawaban D.