Substitution - Clavius

Substitution Method

Example 01

$\int (2 x - 10)^{3} d x$

Let:

$\begin{array}{r} u = 2 x - 10 \end{array}$

$\begin{aligned} \frac{d u}{d x} = 2 \\ d u = 2 d x \\ d x = \frac{d u}{2} \end{aligned}$

$\begin{aligned} \int (x - 2)^{3} d x \\ \int u^{3} \frac{d u}{2} \\ \frac{1}{2} \int u^{3} d u \\ \frac{1}{2} . \frac{1}{4} . u^{4} + c \\ \frac{1}{8} u^{4} + c \\ \frac{1}{8} (2 x - 10)^{4} + c \end{aligned}$

Example 02

$\int (2 x + 5) (x^{2} + 5 x)^{6} d x$

Let:

$\begin{array}{r} u = x^{2} + 5 x \end{array}$

$\begin{aligned} \frac{d u}{d x} = 2 x + 5 \\ d u = (2 x + 5) d x \\ d x = \frac{d u}{(2 x + 5)} \end{aligned}$

$\begin{aligned} \int (2 x + 5) (x^{2} + 5 x)^{6} d x \\ \int (2 x + 5) . u^{6} \frac{d u}{(2 x + 5)} \\ \int (2 x + 5) . u^{6} \frac{d u}{(2 x + 5)} \\ \int u^{6} d u \\ \frac{1}{6 + 1} . u^{6 + 1} + c \\ \frac{1}{7} u^{7} + c \\ \frac{1}{7} (x^{2} + 5 x)^{7} + c \end{aligned}$

Example 03

$\int \frac{1}{3 + \frac{1}{2} x} d x$

$(3 + \frac{1}{2} x)$ will be the new variable

$\begin{aligned} \int \frac{1}{3 + \frac{1}{2} x} d x \\ \int \frac{1}{3 + \frac{1}{2} x} \frac{d (3 + \frac{1}{2} x)}{\frac{1}{2}} \frac{\to (3 + \frac{1}{2} x) as variable}{\to derivative of (3 + \frac{1}{2} x)} \\ 2 \int \frac{1}{3 + \frac{1}{2} x} d (3 + \frac{1}{2} x) \\ 2 \ln (3 + \frac{1}{2} x) + c \\ \ln {(3 + \frac{1}{2} x)}^{2} + c \end{aligned}$

Example 04

$\int 18 . (3 x - 2)^{5} d x$

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

$\begin{aligned} \int 18 . (3 x - 2)^{5} d x \\ \int 18^{6} . (3 x - 2)^{5} \frac{d (3 x - 2)}{3} \frac{\to (3 x - 2) as variable}{\to derivative of (3 x - 2)} \\ \int 6 . (3 x - 2)^{5} d (3 x - 2) \\ 6 . \frac{1}{6} . (3 x - 2)^{6} + c \\ (3 x - 2)^{6} + c \end{aligned}$

Example 05

$\int \frac{d x}{(4 x - 1)^{3}}$

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

$\begin{aligned} \int \frac{d x}{(4 x - 1)^{3}} \\ \int \frac{1}{(4 x - 1)^{3}} \frac{d (4 x - 1)}{4} \frac{\to (4 x - 1) as variable}{\to derivative of (4 x - 1)} \\ \frac{1}{4} \int (4 x - 1)^{- 3} d (4 x - 1) \\ \frac{1}{4} . \frac{1}{- 3 + 1} . (4 x - 1)^{- 3 + 1} + c \\ \frac{1}{4} . \frac{1}{- 2} . (4 x - 1)^{- 2} + c \\ - \frac{1}{8} . \frac{1}{(4 x - 1)^{2}} + c \\ \frac{- 1}{8 (4 x - 1)^{2}} + c \end{aligned}$

Exercise

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Substitution Method

Exercise

Integration