Metode Substitusi 1

Metode substitusi U

\begin{equation*}
\begin{split}
u & = 2x - 10 \\\\
\frac{du}{dx} & = 2 \rightarrow dx & = \frac{du}{2}
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
& \int (x-2)^3\:dx \\\\
& \int u^3 \: \frac{du}{2} \\\\
& \frac{1}{2} \int u^3 \: du \\\\
& \frac{1}{2} \:.\: \frac{1}{4} \:.\: u^4 +C \\\\
& \frac{1}{8} u^4 +C \\\\
& \frac{1}{8} (2x-10)^4 +C
\end{split}
\end{equation*}

Metode substitusi langsung

\begin{equation*}
\begin{split}
& \int(2x-10)^3\:dx \\\\
& \int (2x-10)^3\:\frac{d(2x-10)}{2} \\\\
& \frac{1}{2}\int (2x-10)^3\:d(2x-10) \\\\
& \frac{1}{2} \:.\: \frac{1}{4}(2x-10)^4+C \\\\
& \frac{1}{8}(2x-10)^4+C
\end{split}
\end{equation*}

Metode substitusi U

\begin{equation*}
\begin{split}
u & = x^2 + 5x \\\\
\frac{du}{dx} & = 2x + 5 \rightarrow dx & = \frac{du}{2x + 5}
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
& \int (2x + 5)(x^2 + 5x)^6\:dx \\\\
& \int  (2x + 5) \: u^6 \: \frac{du}{2x + 5} \\\\
& \int  \cancel {(2x + 5)} \: u^6 \: \frac{du}{\cancel {(2x + 5)}} \\\\
& \int u^6 \: du \\\\
& \frac{1}{7}  u^7 +C \\\\
& \frac{1}{7} ((x^2 + 5x)^7 +C
\end{split}
\end{equation*}

Metode langsung

\begin{equation*}
\begin{split}
& \int (2x + 5)(x^2 + 5x)^6\: dx \\\\
& \int (2x + 5)(x^2 + 5x)^6\: \frac{d(x^2 + 5)}{2x + 5x} \\\\
& \int (\cancel {2x + 5})(x^2 + 5x)^6\: \frac{d(x^2 + 5)}{\cancel {2x + 5}} \\\\
& \int (x^2 + 5x)^6\: d(x^2 + 5x) \\\\
& \frac{1}{7}(x^2 + 5x)^7 + C
\end{split}
\end{equation*}

Metode substitusi U

\begin{equation*}
\begin{split}
u & = x + 2 \\\\
\frac{du}{dx} & = 1 \rightarrow dx & = du
\end{split}
\end{equation*}

\begin{equation*}
\begin{split}
& \int x \sqrt{x + 2} \:dx \\\\
& \int  (u - 2) \sqrt{u} \: du \\\\
& \int u\sqrt{u} - 2 \sqrt{u} \: du \\\\
& \int u^{1 \frac 12} - 2 u^{\frac 12} \: du \\\\
& \frac 25 u^{2 \frac 12} - 2 \:.\: \frac 23 u^{1\frac 12} + C \\\\
& \frac 25 u^2 \sqrt{u} - \frac 43 u \sqrt{u} + C\\\\
& \frac {2}{15} u \sqrt{u} (3u - 10) + C\\\\
& \frac {2}{15} (x + 2) \sqrt{x + 2} [3(x + 2) - 10] + C \\\\
& \frac {2}{15} (x + 2) (3x - 4)  \sqrt{x + 2}+ C
\end{split}
\end{equation*}

Metode langsung

\begin{equation*}
\begin{split}
& \int x \sqrt{x + 2} \:dx \\\\
& \int (x + 2 - 2) \sqrt{x + 2} \:dx \\\\
& \int (x + 2) \sqrt{x + 2} - 2 \sqrt{x + 2} \: \frac {d(x + 2)}{1} \\\\
& \int (x + 2)^{1 \frac 12} \:d(x + 2) - \int 2 (x + 2)^{\frac 12} \: d(x + 2) \\\\
& \frac 25 (x + 2)^{2 \frac 12} - 2 \:.\: \frac 23 (x + 2)^{1 \frac 12} + C \\\\
& \frac 25 (x + 2)^2 \sqrt{x + 2} - \frac 43 (x + 2) \sqrt{x + 2} + C \\\\
& \frac {2}{15} (x + 2) \sqrt{x + 2} [3(x + 2) - 10] + C \\\\
& \frac {2}{15} (x + 2) (3x - 4)  \sqrt{x + 2}+ C
\end{split}
\end{equation*}