Integrating (ax + b)ⁿ

Find $\displaystyle\int (3x - 2)^4\,dx$, or $\displaystyle\int \dfrac{1}{(2x + 3)^2}\,dx$, or $\displaystyle\int \sqrt{4x + 1}\,dx$.

Inside the bracket there's always a number times $x$, plus a number: $ax + b$. Expanding $(3x - 2)^4$ would take ages, so here's the one-line way.

1. Write it as a bracket to a power. Roots and fractions too: $\sqrt{4x + 1} = (4x + 1)^{\frac12}$.
2. Add 1 to the power. The bracket stays exactly the same.
3. Divide by the new power.
4. Divide by the number in front of $x$. This is the step people forget.
5. Add $+\,c$.

In symbols:

$\int (ax + b)^n\,dx = \dfrac{(ax + b)^{n+1}}{a(n+1)} + c \qquad (n \neq -1)$

• $a$ is the number in front of $x$ (it can be negative), and $b$ is the number added.
• $n$ is the power. $n = -1$ is excluded because then $n + 1 = 0$, and you can't divide by 0.

In words: power up, divide by the new power, divide by $a$, add $c$.