Integration by parts

The product rule, backwards

Substitution (lesson 2.6.1) undoes the chain rule. Integration by parts undoes the product rule. It turns the integral of a product into a different integral, in which one factor has been differentiated and the other integrated.

Call the two factors $u$ and $v$. The product rule (lesson 2.3.2) says $(uv)' = u'v + uv'$.

Integrate both sides: the left side gives back $uv$, so $uv = \int u'v\,dx + \int uv'\,dx$. Move one integral across and

The shorthand $dv = v'\,dx$ and $du = u'\,dx$, as in lesson 2.6.1, gives the usual form.

A first example: $x\,e^x$

In $\int x\,e^x\,dx$, differentiating $x$ gives 1, which is simpler. Integrating $e^x$ gives $e^x$ again, which is no worse. So choose $u = x$ and $dv = e^x\,dx$.

1. 1
   $$\begin{gathered} u = x, \quad dv = e^x\,dx \\ du = dx, \quad v = e^x \end{gathered}$$

   Why?
2. 2
   $$\begin{aligned} &\int x\,e^x\,dx \\ &= x\,e^x - \int e^x\,dx \end{aligned}$$

   Why?
3. 3
   $$\begin{aligned} &\int x\,e^x\,dx \\ &= x\,e^x - e^x + C \\ &= (x - 1)\,e^x + C \end{aligned}$$

   Why?

Check with the product rule: $\frac{d}{dx}\big[(x - 1)e^x\big] = e^x + (x - 1)e^x = x\,e^x$.

Numbers agree too. With $F(x) = (x - 1)e^x$, the area under $t\,e^t$ from 0 to $x$ is $F(x) - F(0) = (x - 1)e^x + 1$ (lesson 2.5.3). A numerical integral gives the same values.

The area picture

Lesson 2.3.2 drew the product $uv$ as the area of a rectangle. Integration by parts splits that rectangle into two pieces, one for each integral.

Take $\int_1^e \ln x\,dx$ with $u = \ln x$ and $v = x$. As $x$ runs from 1 to $e$, the point $(v, u) = (x, \ln x)$ traces a curve from $(1, 0)$ to $(e, 1)$.

• Under the curve, measured along the $v$ axis, is $\int u\,dv = \int_1^e \ln x\,dx$, the integral we want.
• Beside the curve, measured along the $u$ axis, is $\int v\,du$. Here $u$ runs from 0 to 1 and $v = e^u$, so this area is $\int_0^1 e^u\,du = e - 1$.
• Together they fill the rectangle $e \times 1$. The value $uv$ at the start is $1 \times 0 = 0$, so nothing is cut off there.

So $\int_1^e \ln x\,dx = e - (e - 1) = 1$. In general $[uv]$ is the rectangle at the end minus the rectangle at the start, and the two integrals share what is left.

Choosing the parts

Choose $u$ to be the factor that gets simpler when you differentiate it. Choose $dv$ to be a factor you can integrate, and that does not get worse. Each line of the table leaves out the $+ C$.

• A log goes in $u$. Its derivative $\frac{1}{x}$ is simpler, and it is hard to integrate. For $\ln x$ alone, the other factor is an unseen 1, so $dv = dx$ and $v = x$.
• Otherwise a power of $x$ goes in $u$. Each derivative lowers its power by one.
• Watch the sign of $v$. For $dv = e^{-x}\,dx$, $v = -e^{-x}$: dividing by $k = -1$, as in lesson 2.5.2. So $\int x\,e^{-x}\,dx = -x\,e^{-x} + \int e^{-x}\,dx = -(x + 1)e^{-x} + C$.

The other choice for $x\,e^x$ shows what goes wrong. With $u = e^x$ and $dv = x\,dx$, $v = \frac{x^2}{2}$ and the new integral is $\int \frac{x^2}{2}e^x\,dx$. The power of $x$ went up, so the integral got harder.

Parts with limits

For a definite integral, evaluate $uv$ at both limits, as in the fundamental theorem (lesson 2.5.3).

For $\int_0^1 x\,e^{-x}\,dx$, take $u = x$ and $v = -e^{-x}$:

That is about 0.264241. The figure's integral works the same way: $\int_1^e \ln x\,dx = \big[x\ln x\big]_1^e - \int_1^e 1\,dx = e - (e - 1) = 1$.

Repeated parts

One round of parts lowers a power of $x$ by one. If a power is still left, use parts again.

For $\int x^2 e^x\,dx$, take $u = x^2$ and $dv = e^x\,dx$. Then $\int x^2 e^x\,dx = x^2 e^x - \int 2x\,e^x\,dx$. The new integral is twice the first example, $2(x - 1)e^x$, so

With limits, $\int_0^1 x^2 e^x\,dx = (1 - 2 + 2)e - 2 = e - 2 \approx 0.718282$. A cube needs three rounds: $\int x^3 e^x\,dx = (x^3 - 3x^2 + 6x - 6)e^x + C$.

An integral that comes back

Sometimes parts brings back the integral you started with. Then treat it as an unknown and solve for it, as in lesson 0.3.2.

Call $I = \int \frac{\ln x}{x}\,dx$. Take $u = \ln x$ and $dv = \frac{1}{x}\,dx$, so $du = \frac{1}{x}\,dx$ and $v = \ln x$. Parts gives

Add $I$ to both sides: $2I = (\ln x)^2$, so $I = \frac{(\ln x)^2}{2} + C$. Substitution found the same answer in lesson 2.6.1.

The best-known case is $\int e^x \sin x\,dx$, where two rounds bring it back. It needs the slopes of $\sin x$ and $\cos x$ from lesson 2.3.5: $\cos x$ and $-\sin x$. The answer is $\frac{1}{2}e^x(\sin x - \cos x) + C$.

When parts does not help

Parts needs a factor you can integrate. In $\int x\,e^{x^2}\,dx$, the choice $u = x$ would need $v = \int e^{x^2}\,dx$, which has no formula. Substitution (lesson 2.6.1) does it in one line: $\frac{1}{2}e^{x^2} + C$.

Some integrals defeat both methods. For $\int e^{x^2}\,dx$, parts with $u = e^{x^2}$ and $dv = dx$ gives $x\,e^{x^2} - \int 2x^2 e^{x^2}\,dx$, which is worse. Lesson 2.6.3 finds areas under the related bell curve $e^{-x^2}$ by other means.

A rule of thumb: first look for an inside and its derivative (substitution). If there is none, and the integrand is a power or a log times something you can integrate, try parts. Lesson 2.6.3 uses parts to find the area under $x\,e^{-x}$ out to infinity.

Common mistakes

1. Choose $u$ to get simpler. With $u = e^x$ and $dv = x\,dx$, the new integral $\int \frac{x^2}{2}e^x\,dx$ is harder.
2. Keep the minus sign in $uv - \int v\,du$.
3. Get the sign of $v$ right. For $dv = e^{-x}\,dx$, $v = -e^{-x}$, not $e^{-x}$.
4. Use both limits on $uv$. $\big[uv\big]_a^b$ is the value at $b$ minus the value at $a$, even when the second is not 0.
5. Keep going while a power is left. $\int x^2 e^x\,dx$ needs two rounds.
6. When the integral comes back, solve for it. $I = (\ln x)^2 - I$ gives $2I = (\ln x)^2$.
7. Try substitution first when an inside and its derivative are both there.