# Integration

## Substitution

###### Substitution Method

Example 01

\begin{equation*} \int (2x-10)^3\:dx \end{equation*}

Let:

\begin{equation*} \begin{split} u = 2x - 10 \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{du}{dx} = 2 \\\\ & du = 2 \: dx \\\\ & 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 \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{1}{8} (2x-10)^4 +c} \end{split} \end{equation*}

Example 02

\begin{equation*} \int (2x + 5)(x^2 + 5x)^6\:dx \end{equation*}

Let:

\begin{equation*} \begin{split} u = x^2 + 5x \end{split} \end{equation*}

\begin{equation*} \begin{split} & \frac{du}{dx} = 2x + 5 \\\\ & du = (2x + 5) \: dx \\\\ & 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}{6 + 1} \:.\: u^{6 + 1} +c \\\\ & \frac{1}{7}  u^7 +c \\\\ & \bbox[5px, border: 2px solid magenta] {\frac{1}{7} (x^2 + 5x)^7 +c} \end{split} \end{equation*}

Example 03

\begin{equation*} \int \dfrac {1}{3 + \frac 12 x} \:dx \end{equation*}

$$\left(3 + \frac 12 x \right)$$ will be the new variable

\begin{equation*} \begin{split} & \int \frac {1}{3 + \frac 12 x} \:dx \\\\ & \int \frac {1}{3 + \frac 12 x} \: \frac{d \left(3 + \frac 12 x \right)}{\frac 12} \quad \frac {{\color {red} \rightarrow \left(3 + \frac 12 x \right) \text{ as variable}}}{{\color {red} \rightarrow  \text{derivative of } \left(3 + \frac 12 x \right)}}\\\\ & 2 \int \frac {1}{3 + \frac 12 x} \: d \left(3 + \tfrac 12 x \right) \\\\ & 2 \ln \left(3 + \tfrac 12 x \right) + c \\\\ & \bbox[5px, border: 2px solid magenta] {\ln \left(3 + \tfrac 12 x \right)^2 + c} \end{split} \end{equation*}

Example 04

\begin{equation*} \int 18 \:.\: (3x-2)^5\:dx \end{equation*}

$$(3x - 2)$$ will be the new variable

\begin{equation*} \begin{split} & \int 18 \:.\: (3x-2)^5\:dx \\\\ & \int \cancelto {6} {18}  \:.\: (3x-2)^5\:\frac{d(3x-2)}{\cancel{3}} \quad \frac {{\color {red} \rightarrow (3x - 2) \text{ as variable}}}{{\color {red} \rightarrow  \text{derivative of } (3x - 2)}} \\\\ & \int 6 \:.\: (3x-2)^5\:d(3x-2) \\\\ & 6 \:.\: \frac{1}{6} \:.\: (3x-2)^6 + c \\\\ & \bbox[5px, border: 2px solid magenta] {(3x-2)^6 + c} \end{split} \end{equation*}

Example 05

\begin{equation*} \int \dfrac {dx}{(4x - 1)^3} \end{equation*}

$$(4x - 1)$$ akan dijadikan sebagai variabel

\begin{equation*} \begin{split} & \int \frac {dx}{(4x - 1)^3} \\\\ & \int \frac {1}{(4x - 1)^3} \: \frac{d(4x - 1)}{4} \quad \frac {{\color {red} \rightarrow (4x - 1) \text{ as variable}}}{{\color {red} \rightarrow  \text{derivative of } (4x - 1)}}\\\\ & \frac 14 \int (4x - 1)^{-3} \: d(4x - 1) \\\\ & \frac 14 \:.\: \frac {1}{-3 + 1} \:.\: (4x - 1)^{-3 + 1} + c \\\\ & \frac 14 \:.\: \frac {1}{-2} \:.\: (4x - 1)^{-2} + c \\\\ & - \frac 18 \:.\: \frac {1}{(4x - 1)^2} + c \\\\ & \bbox[5px, border: 2px solid magenta] {\frac {-1}{8 (4x - 1)^2} + c} \end{split} \end{equation*}