Question regarding "integration by parts"

Question

I'm reading about "integration by parts", and I have a hard time understanding why in the below example they do the procedure 3 times. Or maybe better put, when do you know that you have to do integration by parts more in order to be finished? Function to be integrated Step by step how the book does it

Can't you just simply say that you are finish integrating, when we obtain the integral for the last "remainder" part, which in this example was xe^2x in part **?

Answer 1

The answer to "When do you stop integrating?" is "When you have something you know how to integrate".

Do you, inherently, know what $\int x^3 e^x dx$ is? It doesn't normally appear on a table of standard integrals. We do know how to find $\int x^3 dx$ and $\int e^x dx$, though, and so the aim of using IBP is to hopefully make use of these integrals we do know somehow.

If we apply IBP in the direction shown in the textbook, we find that we're left with a term like $\int x^2 e^x dx$, and while it still isn't something we can read off a table, we notice that we've reduced the $x^3$ to $x^2$ and intuitively we can see that if we repeat the process we're likely to keep reducing the exponent of $x$ - from $x^3$ to $x^2$ to $x$ to just a constant, and once we do that we know we'll be left with just the nice simple integral $\int e^x dx$.

Now we could apply IBP the other way around - treating $x^3$ as the $g'$ and $e^x$ as the $f$. But what we find if we do that is that we end up increasing the exponent of the polynomial term - we go from $x^3$ to $x^4$, and if we continue in that direction we're just going to get increasingly more complicated integrals that we don't know how to approach.

A lot of integration is like this - applying techniques like substitution and integration by parts to transform an ugly integral into something more familiar. Sometimes, it takes a mix of experience and trial-and-error to get to the answer.

Answer 2

Assume that we can differentiate $f$, and we can integrate $g$. Integration by parts exchanges the problem of computing

$$\int fg dx\tag 1$$

with that of computing

$$-\int f'G dx\tag 2$$

plus an extra term, where $G$ is an antiderivative of $g$. This can be repeated to get to

$$(-1)^n\int f^{(n)}g^{(-n)} dx\tag 3$$

again plus some other terms, where $f^{(n)}$ is the $n$-fold derivative of $f$, and $g^{(-n)}$ is the $n$-fold antiderivative of $g$.

So, applying this rule $n$-times is useful precisely when $(3)$ happens to be easier than $(1)$.

Some cases when this happens:

1. If $f$ is a polynomial, then $f^{(n)}=0$ for some $n$, and $(3)$ vanishes and is therefore easier than $(1)$.
2. If $f$ satisfies a differential equation, e.g. $f=\sin x$, then $(1)$ and $(3)$ may be related.

The example $\int x^3e^{2x}dx$ is one where $f=x^3$ is a polynomial, thus integrating by parts 4 times will give zero, so the whole answer is just the `extra terms' I glossed over above. Moreover, $g=e^{2x}$ is easily integrable, which is another requirement for this to work. In situation 1. the process can be done very quickly as is explained here.

Answer 3

Whenever you have an power and a transcendental function (namely a trigonometric/exponential function like $\sin x$ or $e^x$), each time you apply IBP, the power in $x$ reduces, and you will get a simpler function in $x$ until you reach a constant, leaving you with a simple transcendental function.

For your case, $x^3$ can be differentiated to $3x^2$, then $6x$, then $6$ and then zero, while $e^{2x}$ will integrate to $\frac {1}{2} e^{2x}$ each time. Thus, you have to carry out IBP several times and add your prior results until you get to the simplest function without a power in $x$, usually the constant.

\begin{matrix} u = x^3 & dv = e^{2x} \\ du = 3x^2 & v = \frac {1}{2} e^{2x}\\ \end{matrix}

$$\therefore uv - \int v \ du = \frac {1}{2} x^3 e^{2x} - \int 3x^2 e^{2x} \ dx$$

\begin{matrix} u = 3x^2 & dv = e^{2x} \\ du = 6x & v = \frac {1}{2} e^{2x}\\ \end{matrix}

$$\therefore uv - \int v \ du = \frac {1}{2} x^3 e^{2x} - \frac {3}{2} x^2 e^{2x} - \int 3x e^{2x} \ dx$$

\begin{matrix} u = 3x & dv = e^{2x} \\ du = 3 & v = \frac {1}{2} e^{2x}\\ \end{matrix}

$$\therefore uv - \int v \ du = \frac {1}{2} x^3 e^{2x} - \frac {3}{2} x^2 e^{2x} + \frac {3}{2} e^{2x}- \int \frac {3}{2} e^{2x} \ dx$$

Since the final integral is simple ($\int \frac {3}{2} e^{2x} \ dx$) it'll be $\frac {3}{4} e^{2x}$, so our complete function is $$\frac {1}{2} x^3 e^{2x} - \frac {3}{2} x^2 e^{2x} + \frac {3}{2} x e^{2x} - \frac {3}{4} e^{2x} + C$$ and if we wanted to clean it up we'd get $$\frac {1}{2} e^{2x} (x^3 -3x^2 +3x -3/2) + C$$