∫x√(3x+5) dx
Selesaikan bentuk nya…
[tex]\displaystyle\int x\sqrt{3x+5}\,dx[/tex]
[tex]\displaystyle=\bf\frac{1}{9}\left(\frac{2}{5}(3x+5)^{{}^5\!/_2}-\frac{10}{3}(3x+5)^{{}^3\!/_2}\right)+C[/tex]
atau
[tex]\displaystyle\int x\sqrt{3x+5}\,dx[/tex]
[tex]\displaystyle=\bf\frac{1}{9}\left(\frac{2}{5}\sqrt{{(3x+5)}^5}-\frac{10}{3}\sqrt{{(3x+5)}^3}\right)+C[/tex]
Pembahasan
Integral
[tex]\begin{aligned}&\int x\sqrt{3x+5}\,dx\\&{=\ }\int x\sqrt{u}\,dx\quad\left(\textsf{dengan $u=3x+5$}\right)\\&\quad\left[\ \begin{aligned}&u=3x+5\\&\Rightarrow \frac{du}{dx}=3\\&\Rightarrow dx=\frac{1}{3}\,du\\&\Rightarrow x=\frac{u-5}{3}\end{aligned}\right.\\&{=\ }\int\frac{u-5}{3}\sqrt{u}\cdot\frac{1}{3}\,du\\&{=\ }\frac{1}{9}\int\sqrt{u}(u-5)\,du\\&{=\ }\frac{1}{9}\int\left(u\sqrt{u}-5\sqrt{u}\right)du\\&{=\ }\frac{1}{9}\int\left(u^{{}^3\!/_2}-5u^{{}^1\!/_2}\right)du\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\frac{1}{9}\left(\int u^{{}^3\!/_2}\,du-5\int u^{{}^1\!/_2}\,du\right)\\&{=\ }\frac{1}{9}\left(\frac{u^{\left(1+{}^3\!/_2\right)}}{1+{}^3\!/_2}-\frac{5u^{\left(1+{}^1\!/_2\right)}}{1+{}^1\!/_2}\right)\\&{=\ }\frac{1}{9}\left(\frac{u^{{}^5\!/_2}}{{}^5\!/_2}-\frac{5u^{{}^3\!/_2}}{{}^3\!/_2}\right)\\&{=\ }\frac{1}{9}\left(\frac{2}{5}u^{{}^5\!/_2}-\frac{10}{3}u^{{}^3\!/_2}\right)\end{aligned}[/tex]
[tex]\begin{aligned}&{=\ }\boxed{\ \bf\frac{1}{9}\left(\frac{2}{5}(3x+5)^{{}^5\!/_2}-\frac{10}{3}(3x+5)^{{}^3\!/_2}\right)+C\ }\\&{=\ }\boxed{\ \bf\frac{1}{9}\left(\frac{2}{5}\sqrt{{(3x+5)}^5}-\frac{10}{3}\sqrt{{(3x+5)}^3}\right)+C\ }\end{aligned}[/tex]
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