Further Integration of Brackets

Video lesson · from 59:332 worked examples

One lesson video covers all of Integration, so it opens at 59:33 for this topic — not from the beginning.

We have previously learned how to differentiate functions of the form $f(ax+b)$ :

\frac{d}{dx}[(ax+b)^{n+1}] = (n+1)(ax+b)^n \times a = a(n+1)(ax+b)^n

For integrating linear brackets raised to a power, we use:

\int (ax+b)^n \,dx = \frac{(ax+b)^{n+1}}{a(n+1)} + C \quad (n \neq -1)

Divide by the derivative of the bracket: integrating $(3x+2)^4$ needs $\times \frac{1}{3}$ as well as raising the power. Forgetting that factor is the standard error.
This only works for a linear bracket: the shortcut is valid when the inside is $ax+b$ . It does not work when the bracket contains $x^2$ .
Add one to the power, then divide by the new power: and by the coefficient — two divisions, not one.
Rewrite fractions as negative powers first: $\frac{1}{(2x-1)^2}$ becomes $(2x-1)^{-2}$ .
Still write $+\,C$ : the bracket form makes it especially easy to forget.

Example 1

Find $\int (5x+2)^2 \, dx$ .

\begin{aligned} \int (5x+2)^2 \, dx &= \frac{(5x+2)^3}{5 \times 3} + C \\ &= \frac{(5x+2)^3}{15} + C \end{aligned}

Example 2

Find $\int \frac{1}{\sqrt[3]{5x+2}} \, dx$ .

First, rewrite the expression with a fractional exponent:

\int (5x+2)^{-1/3} \, dx

Now integrate using the rule:

\begin{aligned} &= \frac{(5x+2)^{2/3}}{5 \times \frac{2}{3}} + C \\ &= \frac{3(5x+2)^{2/3}}{10} + C \\ &= \frac{3\sqrt[3]{(5x+2)^2}}{10} + C \end{aligned}