적분 전략 고르기

If the unknown exponent gives you the jitters, think about what this looks like in easier cases. If $n$ is a whole number, the integrand is a polynomial. Not so scary, right? However, it's a little complicated to expand. (You can do it using the very handy binomial theorem.) Let's think of an easier way.

If we had simply the variable $x$ raised to the power $n$, rather than the binomial $x+a$, that might be nicer. So, let's use the substitution $u=x+a$, $\mathrm{d}u=\mathrm{d}x$. Note $x=u-a$.

$$\begin{aligned} \int x(x+a)^n~\mathrm{d}x&=\int (u-a)u^n~\mathrm{d}x = \int \big(u^{n+1}-au^n\big)~\mathrm{d}u \\ \text{Now, if $n\neq -1$ and $n \neq -2$, we can just use the power rule:} \\ &=\frac{u^{(n+2)}}{n+2}-a\frac{u^{n+1}}{n+1}+C\\ &=\frac{(x+a)^{(n+2)}}{n+2}-a\frac{(x+a)^{n+1}}{n+1}+C \\ \text{If $n=-1$, then} \\ \int x(x+a)^n~\mathrm{d}x&=\int \big(u^{n+1}-au^n\big)~\mathrm{d}u =\int \left(1 - \frac{a}{u}\right)~\mathrm{d}u \\&= u - a\log|u|+C = (x+a)-a\log|x+a|+C \\ \text{If $n=-2$, then} \\ \int x(x+a)^n~\mathrm{d}x&=\int \big(u^{n+1}-au^n\big)~\mathrm{d}u=\int\left( \frac{1}{u} - au^{-2}\right)~\mathrm{d}u \\& = \log|u| + \frac{a}{u}+C= \log|x+a| + \frac{a}{x+a}+C \\ \text{All together,} \\ \int x(x+a)^n~\mathrm{d}x&=\begin{cases} \frac{(x+a)^{(n+2)}}{n+2}-a\frac{(x+a)^{n+1}}{n+1}+C&\text{ if } n \neq -1,-2\\ (x+a)-a\log|a+x|+C & \text{ if } n=-1\\ \log|x+a| +\frac{a}{x+a}+C& \text{ if } n=-2 \end{cases} \end{aligned}$$