적분 기법 전체 문제

The substitution $u=x+1$ looks promising at first, but doesn't result in something easily integrable. We can't use partial fractions because our integration isn't rational. This doesn't look like something from the trig-substitution family. So, let's think about integration by parts. There's a lot of different ways we could break up the integrand into two parts. For example, we could view it as $\Big(\frac{x}{(x+1)^2}\Big)\Big(e^x\Big)$, or we could view it as $\Big(\frac{x}{x+1}\Big)\Big(\frac{e^x}{x+1}\Big)$. After some trial and error, we settle on $u=xe^x$ and $\mathrm{d}v=(x+1)^{-2}~\mathrm{d}x$. Then $\mathrm{d}u=e^x(x+1)$ and $v=\frac{-1}{x+1}$.

$$\begin{aligned} \int\frac{xe^x}{(x+1)^2}~\mathrm{d}x&=-\frac{xe^x}{x+1} +\int \frac{e^x(x+1)}{x+1}\mathrm{d}x \\&=-\frac{xe^x}{x+1} +\int e^x\mathrm{d}x\\ &=-\frac{xe^x}{x+1}+e^x+C \\&= \frac{e^x}{x+1}+C \end{aligned}$$

It would be nice to use integration by parts with $u=x$, because then we would integrate $\int v~\mathrm{d}u$, and $\mathrm{d}u = \mathrm{d}x$. That is, the $x$ would go away, and we'd be left with a pure trig integral. If we use $u=x$, then $\mathrm{d}v = \frac{\sin x}{\cos^2 x}$. We need to find $v$:

$$\begin{aligned} v&=\int \frac{\sin x}{\cos^2 x}~\mathrm{d}x = \int \tan x \sec x~\mathrm{d}x = \sec x \\ \text{Now we use integration by parts.} \\ \int \frac{x\sin x}{\cos^2 x}\mathrm{d}x&=x\sec x - \int \sec x ~\mathrm{d}x= x\sec x - \log|\sec x + \tan x|+C \end{aligned}$$

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 $x=u-a$.

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