Substitution for (ax+b)

Setting $t=ax+b$ gives $dt=a\,dx$, that is, $dx=\frac{1}{a}\,dt$. This $\frac{1}{a}$ comes out in front as a coefficient.

When you differentiate the answer, the chain rule multiplies by the derivative of the inside, $a$. The $\frac{1}{a}$ is there to cancel it. When $a=1$, $\frac{1}{a}=1$, and the formula looks the same as the one for $x^n$.

Examples

1. 01Integrate$\displaystyle \int \left(3x + 2\right)^{3} \, dx$AnswerHide

   Answer$\displaystyle \frac{\left(3x + 2\right)^{4}}{12} + C$
2. 02Integrate$\displaystyle \int \left(3x + 2\right)^{2} \, dx$AnswerHide

   Answer$\displaystyle \frac{\left(3x + 2\right)^{3}}{9} + C$
3. 03Integrate$\displaystyle \int \left(x + 2\right)^{3} \, dx$AnswerHide

   Answer$\displaystyle \frac{\left(x + 2\right)^{4}}{4} + C$
4. 04Integrate$\displaystyle \int \left(x + 3\right)^{4} \, dx$AnswerHide

   Answer$\displaystyle \frac{\left(x + 3\right)^{5}}{5} + C$
5. 05Integrate$\displaystyle \int \left(2x + 2\right)^{3} \, dx$AnswerHide

   Answer$\displaystyle \frac{\left(2x + 2\right)^{4}}{8} + C$
6. 06Integrate$\displaystyle \int \left(x + 1\right)^{2} \, dx$AnswerHide

   Answer$\displaystyle \frac{\left(x + 1\right)^{3}}{3} + C$

FAQ

Q1Why does $\frac{1}{a}$ appear?AnswerHide

A

Differentiate the answer to check: $\left\{\frac{(3x+2)^{4}}{12}\right\}'=\frac{4(3x+2)^{3}\cdot 3}{12}=(3x+2)^{3}$. The derivative of the inside produces a factor of $3$, so you multiply by $\frac{1}{3}$ in advance.

Q2What about $n=-1$?AnswerHide

A

This formula does not apply. Instead, $\int\frac{dx}{ax+b}=\frac{1}{a}\log|ax+b|+C$. The $\frac{1}{a}$ still appears.