Contoh Soal 3 Materi Integral Tak Tentu Fungsi Aljabar

 $\begin{array}{ll}\\ 11.&\displaystyle \int d\left ( 2x \right )=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle x+C\\ \textrm{b}.\quad \color{red}\displaystyle 2x+C\\ \textrm{c}.\quad \displaystyle 3x+C\\ \textrm{d}.\quad \displaystyle 4x+C\\ \textrm{e}.\quad \displaystyle 5x+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int d\left ( 2x \right )=\int 2\: \: dx=2x+C. \end{array}$.

$\begin{array}{ll}\\ 12.&\displaystyle \int 5\: \: d\left ( 2x \right )=\: ....\\ &\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle 5+C\\ \textrm{b}.\quad \displaystyle \frac{5}{2}x+C\\ \textrm{c}.\quad \displaystyle 5x+C\\ \textrm{d}.\quad \color{red}\displaystyle 10x+C\\ \textrm{e}.\quad \displaystyle 25x+C\end{array}\\\\ &\textbf{Jawab}:\\ &\displaystyle \int 5\: \: d\left ( 2x \right )=\int5. 2\: \: dx=\int 10\: dx=10x+C. \end{array}$.

$\begin{array}{ll}\\ 13.&\displaystyle \int 6x\: \: d\left ( 3x \right )=\: ....\\ &\begin{array}{ll}\\  \textrm{a}.\quad \displaystyle 18x^{2}+C\\ \textrm{b}.\quad \color{red}\displaystyle 9x^{2}+C\\ \textrm{c}.\quad \displaystyle 5x^{2}+C\\ \textrm{d}.\quad \displaystyle 3x^{2}+C\\ \textrm{e}.\quad \displaystyle 2x^{2}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int 6x\: \: d\left ( 3x \right )=\int 6x. 3\: \: dx=\int 18x\: dx\\ &=\displaystyle \frac{18x^{2}}{2}+C=9x^{2}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 14.&\displaystyle \int \left ( 8x^{3}-2x \right )\: \: d\left ( -2x \right )=\: ....\\&\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle -16x^{4}+4x^{2}+C\\ \textrm{b}.\quad \displaystyle 16x^{4}-4x^{2}+C\\ \textrm{c}.\quad \displaystyle -4x^{4}-2x^{2}+C\\ \textrm{d}.\quad \color{red}\displaystyle -4x^{4}+2x^{2}+C\\ \textrm{e}.\quad \displaystyle 4x^{4}-2x^{2}+C\end{array}\\\\ &\textbf{Jawab}:\\ &\begin{aligned}&\displaystyle \int \left (8x^{3}-2x \right )\: \: d\left ( -2x \right )\\ &=\int \left ( 8x^{3}-2x \right ).\left (-2\: \: dx \right )\\&=\int \left (-16x^{3}+4x \right )\: dx\\ &=\displaystyle -\frac{16x^{4}}{4}+\frac{4x^{2}}{2}+C\\ &=\displaystyle -4x^{4}+2x^{2}+C \end{aligned} \end{array}$.

$\begin{array}{ll}\\ 15.&\displaystyle \int \left ( x+3 \right )\: \: d\left ( 2x+6 \right )=\: ....\\&\begin{array}{ll}\\ \textrm{a}.\quad \displaystyle 4x^{2}+24x+C\\ \textrm{b}.\quad \displaystyle 2x^{2}+12x+C\\ \textrm{c}.\quad \color{red}\displaystyle x^{2}+6x+C\\ \textrm{d}.\quad \displaystyle 4x^{2}+C\\ \textrm{e}.\quad \displaystyle x^{2}+C\end{array}\\\\ &\textrm{Jawab}:\\ &\begin{array}{|c|c|}\hline \color{blue}\textrm{Pertama}&\color{red}\textrm{Kedua}\\\hline \begin{aligned}&\displaystyle \int \left ( x+3 \right )\: d\left ( 2x+6 \right )\\ &=\displaystyle \frac{1}{2}\int \left ( 2x+6 \right )\: d\left ( 2x+6 \right )\\ &=\displaystyle \frac{1}{2}\left [ \frac{1}{2}\left ( 2x+6 \right )^{2} \right ]+C\\ &=\displaystyle \frac{1}{4}\left ( 4x^{2}+24x+36 \right )+C\\ &=\displaystyle \frac{4x^{2}}{4}+\frac{24x}{4}+\frac{36}{4}+C\\ &=x^{2}+6x+\underset{C}{\underbrace{9+C}}\\ &=x^{2}+6x+C \end{aligned}&\begin{aligned}&\displaystyle \int \left ( x+3 \right )\: d\left ( 2x+6 \right )\\ &=\int \left ( x+3 \right ).\left ( 2\: dx \right )\\ &=\int \left ( 2x+6 \right )\: dx\\ &=\displaystyle \frac{2x^{2}}{2}+6x+C\\ &=x^{2}+6x+C\\ &\\ &\\ & \end{aligned}\\\hline \end{array} \end{array}$ .