3.2 Integration by Substitution

3.2 Integration by Substitution

It is given that

${\int (a x + b)}^{n} d x, n \neq - 1.$

(A) Using the Substitution method,

$\begin{array}{l} Let u = a x + b \\ Thus, \frac{d u}{d x} = a \\ ∴ d x = \frac{d u}{a} \end{array}$

Example 1:

$\begin{array}{l} {\int (3 x + 5)}^{3} d x . \\ Let u = 3 x + 5 \\ \frac{d u}{d x} = 3 \\ d x = \frac{d u}{3} \\ {\int (3 x + 5)}^{3} d x \\ = \int u^{3} \frac{d u}{3} \leftarrow \begin{array}{l} substitute 3 x + 5 = u \\ and d x = \frac{d u}{3} \end{array} \\ = \frac{1}{3} \int u^{3} d u \\ = \frac{1}{3} (\frac{u^{4}}{4}) + c \\ = \frac{1}{3} (\frac{{(3 x + 5)}^{4}}{4}) + c \leftarrow \begin{array}{l} substitute back \\ u = 3 x + 5 \end{array} \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}$

(B) Using Formula method

$\begin{array}{l} \int {(a x + b)}^{n} = \frac{{(a x + b)}^{n + 1}}{(n + 1) a} + c \\ Hence, \\ \int {(3 x + 5)}^{3} d x = \frac{{(3 x + 5)}^{4}}{4 (3)} + c \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}$

Example 2 (Formula method):

$\begin{array}{l} \int {(a x + b)}^{n} = \frac{{(a x + b)}^{n + 1}}{(n + 1) a} + c \\ Hence, \\ \int {(3 x + 5)}^{3} d x = \frac{{(3 x + 5)}^{4}}{4 (3)} + c \\ = \frac{{(3 x + 5)}^{4}}{12} + c \end{array}$