$\begin{array}{ll}\\ 16.&\textbf{(UN 2015 Matematika IPA)}\\ &\textrm{Hasil}\: \: \displaystyle \int 6x\left ( 1-x^{2} \right )^{4}\: \: dx\: \: \textrm{adalah ....}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \frac{3}{5}\left ( 1+x^{2} \right )^{5}+C\\\\ \textrm{b}.\quad \displaystyle \frac{2}{5}\left ( 1+x^{2} \right )^{5}+C\\\\ \textrm{c}.\quad \displaystyle -\frac{1}{5}\left ( 1-x^{2} \right )^{5}+C\\\\ \textrm{d}.\quad -\displaystyle \frac{2}{5}\left ( 1-x^{2} \right )^{5}+C\\\\ \textrm{e}.\quad \color{red}-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C\end{array}\\\\&\textbf{Jawab}:\\&\color{blue}\textrm{Alternatif 1}\\ &\textrm{Dengan integral substitusi}\\ &\begin{array}{l}\\ \begin{aligned}\displaystyle &\int 6x\left ( 1-x^{2} \right )^{4}\: dx\\ &=\int \left ( 1-x^{2} \right )^{4}.6x\: dx\\ &\begin{cases} u& =1-x^{2} \\ & \\ du& =-2x\: \: \: dx\quad \Rightarrow \quad -3\: dx=6x\: dx \end{cases}\\ &\displaystyle \int u^{4}.\left ( -3\: du \right )\\ &=-3\displaystyle \int u^{4}\: du\\ &=-\displaystyle \frac{3}{5}u^{5}+C\\ &=-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C \end{aligned} \end{array}\\ &\color{blue}\textrm{Alternatif 2}\\ &\begin{aligned} &\displaystyle \int \left ( 1-x^{2} \right )^{4}\: 6x\: dx\\ &=\int \left ( 1-x^{2} \right )^{2}.(-3).\left ( -2x\: dx \right )\\ &=-3\displaystyle \int \left ( 1-x^{2} \right )^{4}.\: \left ( -2x\: dx \right )\\ &=-3\displaystyle \int \left ( 1-x^{2} \right )^{4}\: .d\left ( 1-x^{2} \right )\\ &=-3\left [ \displaystyle \frac{\left ( 1-x^{2} \right )^{5}}{5} \right ]+C\\ &=-\displaystyle \frac{3}{5}\left ( 1-x^{2} \right )^{5}+C\end{aligned} \end{array}$.
$\begin{array}{ll}\\ 17.&\begin{aligned}\displaystyle &\int \frac{dx}{2022x+2023}=\: ....\end{aligned}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{b}.\quad \displaystyle \frac{1}{2021}\ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{c}.\quad \color{red}\displaystyle \frac{1}{2022}\ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{d}.\quad \displaystyle \frac{1}{2023}\ln \left | 2022x+2023 \right |+C\\ &\\ \textrm{e}.\quad \displaystyle \frac{2022}{2023}\ln \left | 2022x+2023 \right |+C \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}&\displaystyle \int \frac{dx}{2022x+2023}\\ &\textrm{Misalkan}\\ &u=2022x+2023\\ &\Leftrightarrow du=2022\: dx\Leftrightarrow \displaystyle \frac{1}{2022}\: du=\: dx\\ &\textrm{Sehingga}\\ &=\displaystyle \int \frac{1}{u}.\left ( \frac{1}{2022}\: du \right )=\displaystyle \frac{1}{2022}\int \frac{1}{u}\: du\\ &=\displaystyle \frac{1}{2022}\ln \left | 2022x+2023 \right |+C \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 18.&\textrm{Diketahui}\: \: \displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\: \: \textrm{jika} \: f(4)=19,\\ &\textrm{ maka}\: \: f(1)=....\\ &\begin{array}{ll}\\ \textrm{a}.\quad 2\\ \textrm{b}.\quad 3\\ \textrm{c}.\quad 4\\ \textrm{d}.\quad \color{red}5\\ \textrm{e}.\quad 6\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\textrm{Perlu diingat bahwa}\: \: \displaystyle \frac{d}{dx}f(x)=h(x)\\ &\Leftrightarrow d(f(x))=h(x)\: dx\\ &\Leftrightarrow \displaystyle \int d(f(x))=\displaystyle \int h(x)\: dx\\ &\textrm{Perhatikan kasus di pada soal di atas}\\ &\displaystyle \frac{d}{dx}f(x)=3\sqrt{x}\Leftrightarrow d\left ( f(x) \right )=3\sqrt{x}\: dx\\ &\Leftrightarrow \displaystyle \int d\left ( f(x) \right )=\int 3\sqrt{x}\: dx=\int 3x^{.^{\frac{1}{2}}}\: dx\\&\Leftrightarrow f(x)=2x^{.^{\frac{3}{2}}}+C=2x\sqrt{x}+C \end{aligned}\\&\begin{aligned}&\textrm{Karena diketahui}\\ &f(4)=19,\: \: \textrm{maka}\\ &\Leftrightarrow 2(4)(\sqrt{4})+C=19\\ &\Leftrightarrow 16+C=19\\ &\Leftrightarrow C=3\\ &\textrm{Sehingga}\quad f(x)=2x\sqrt{x}+3\\ &\textrm{maka}\quad f(1)=2.1.\sqrt{1}+3=2+3=5 \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 19.&\begin{aligned}\displaystyle &-\int \frac{dx}{x^{2}+3x+2}\: \: dx=\: .... \end{aligned}\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle \ln \left | \displaystyle \frac{x+1}{x+2} \right |+C\\ &\\ \textrm{b}.\quad \color{red}\displaystyle \ln \left | \displaystyle \frac{x+2}{x+1} \right |+C\\ &\\ \textrm{c}.\quad \displaystyle \ln \left | x^{2}+3x+2 \right |+C\\ &\\ \textrm{d}.\quad \arctan 2\left ( \displaystyle x+\frac{3}{2} \right )+C\\ &\\ \textrm{e}.\quad -\displaystyle \arctan 2\left (\displaystyle x+ \frac{3}{2}\right )+C \end{array}\\\\&\textbf{Jawab}:\\&\begin{aligned}\displaystyle -\int \frac{dx}{x^{2}+3x+2}&=-\displaystyle \int \frac{1}{\left ( x+1 \right )\left ( x+2 \right )}\: dx\\ &=-\displaystyle \int \left ( \displaystyle \frac{1}{x+1}-\displaystyle \frac{1}{x+2} \right )\: dx\\ &=-\displaystyle \int \frac{1}{x+1}\: dx+\displaystyle \int \frac{1}{x+2}\: dx\\ &=\displaystyle \int \frac{1}{x+2}\: dx-\displaystyle \int \frac{1}{x+1}\: dx\\ &=\displaystyle \ln \left | x+2 \right |-\displaystyle \ln \left | x+1 \right |+C\\ &=\displaystyle \ln \left | \displaystyle \frac{x+2}{x+1} \right |+C \end{aligned} \end{array}$.
$\begin{array}{ll}\\ 20.&\textrm{Tentukanah integral berikut}?\\ &\textrm{a}.\quad \int dt\\&\textrm{b}.\quad \int 4\: dw\\ &\textrm{c}.\quad \int \left ( x^{3}+5 \right )\: dx\\ &\textrm{d}.\quad \int \left ( x^{\frac{3}{2}}-2\sqrt{x}+1 \right )dx\\ &\textrm{e}.\quad \int \left ( 5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}} \right )dx\\\\ &\textbf{Jawab}:\\ &\begin{array}{l}\\ \begin{aligned}\textrm{a}.\quad \int dt=t+c \end{aligned}\\ \begin{aligned}\textrm{b}.\quad \int 4\: dw=4w+C \end{aligned}\\ \begin{aligned}\textrm{c}.\quad \int (x^{3}+5)\: dx&=\displaystyle \frac{1}{3+1}x^{3+1}+5x+C\\ &=\displaystyle \frac{1}{4}x^{4}+5x+C \end{aligned} \end{array}\\ &\begin{aligned}\textrm{d}.\quad &\int (x^{\frac{3}{2}}+2\sqrt{x}+1)\: dx\\ &=\int \left (x^{\frac{3}{2}}+2x^{\frac{1}{2}}+1 \right )\: dx\\ &=\displaystyle \frac{1}{\displaystyle \frac{3}{2}+1}x^{\frac{3}{2}+1}+\frac{2}{\displaystyle \frac{1}{2}+1}x^{\frac{1}{2}+1} +x+C\\ &=\displaystyle \frac{1}{\displaystyle \frac{5}{2}}x^{\frac{5}{2}}+\frac{2}{\displaystyle \frac{3}{2}}x^{\frac{3}{2}}+x+C\\ &=\frac{2}{5}x^{\frac{5}{2}}+\frac{4}{3}x^{\frac{3}{2}}+x+C\\ &\quad \textrm{atau dapat juga kita menyatakan dengan}\\ &=\frac{2}{5}x^{2\frac{1}{2}}+\frac{4}{3}x^{1\frac{1}{2}}+x+C\\ &=\frac{2}{5}x^{2}\sqrt{x}+\frac{4}{3}x\sqrt{x}+x+C \end{aligned} \\ &\begin{aligned}\textrm{e}.\quad &\int \left ( 5x^{\frac{3}{2}}-2x+3x^{-\frac{1}{3}} \right )dx\\ &=.....\\ &\quad \textrm{silahkan dicoba sendiri sebagai latihan}\end{aligned} \end{array}$.
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