Integration by Parts

Substitution undoes the chain rule. But what about $\int x e^{2x}\,dx$ or $\int \ln x\,dx$? There is no inside function whose derivative is sitting there. These integrals come from the product rule, and integration by parts is the product rule run backwards. In AP Calculus it’s a BC-only technique (in IB it’s AA HL), and it shows up on exams both on its own and inside bigger problems (improper integrals, Taylor series, area and volume).

Key ideas

The formula

The product rule says $\dfrac{d}{dx}(uv) = u\,\dfrac{dv}{dx} + v\,\dfrac{du}{dx}$. Integrate both sides and rearrange:

$$\int u\,dv = uv - \int v\,du$$

You split the integrand into two pieces: $u$ (which you will differentiate) and $dv$ (which you will integrate). The goal is a new integral $\int v\,du$ that is easier than the one you started with.

A tidy way to organize your work is a small box:

	Differentiate	Integrate
Choose	$u = \ldots$	$dv = \ldots$
Find	$du = \ldots$	$v = \ldots$

When you find $v$, leave off the $+ C$. One constant at the very end is enough.

Choosing u: LIATE

Pick $u$ to be the factor that gets simpler when you differentiate it, and $dv$ to be something you can actually integrate. The guide LIATE lists good choices for $u$, best first:

L	I	A	T	E
Logarithms	Inverse trig	Algebraic (powers of $x$)	Trig	Exponentials

For example, in $\int x e^{2x}\,dx$ the $x$ (Algebraic) comes before $e^{2x}$ (Exponential), so $u = x$. LIATE is a guide, not a law: if your choice makes the new integral worse, switch.

Repeated parts and the tabular method

For $\int x^2 e^x\,dx$, one round of parts lowers $x^2$ to $2x$, and a second round lowers it to a constant. Just do parts twice, carefully.

When $u$ is a polynomial that eventually differentiates to $0$, the tabular method (optional, but quick) keeps the bookkeeping straight. Differentiate $u$ down one column until you reach $0$, integrate $dv$ down the other, then multiply along the diagonals with alternating signs $+, -, +, \ldots$ (see Example 3).

Definite integrals by parts

Evaluate the $uv$ part at the limits too:

$$\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du$$

Integrals that come back

For $\int e^x \cos x\,dx$, neither factor ever gets simpler. Do parts twice, and the original integral shows up again on the right side. Treat it as an unknown $I$ and solve for it algebraically (Example 4).

Trig on this page is in radians, as it is throughout calculus. (Grade 11 trig used degrees; the derivative rules for $\sin$ and $\cos$ only work in radians.)

Worked examples

Example 1: A power times an exponential

Find $\displaystyle\int x e^{2x}\,dx$.

Solution. LIATE puts Algebraic before Exponential, so:

	Differentiate	Integrate
Choose	$u = x$	$dv = e^{2x}\,dx$
Find	$du = dx$	$v = \tfrac{1}{2}e^{2x}$

$$\begin{aligned} \int x e^{2x}\,dx &= x \cdot \tfrac{1}{2}e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx \\ &= \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + C \end{aligned}$$

Check: $\dfrac{d}{dx}\left(\tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x}\right) = \tfrac{1}{2}e^{2x} + x e^{2x} - \tfrac{1}{2}e^{2x} = x e^{2x}$. ✓

Example 2: The integral of ln x

Find $\displaystyle\int \ln x\,dx$, then evaluate $\displaystyle\int_1^e \ln x\,dx$.

Solution. There is only one factor, so use the trick of letting $dv = dx$. Logarithms are first in LIATE, so $u = \ln x$:

	Differentiate	Integrate
Choose	$u = \ln x$	$dv = dx$
Find	$du = \tfrac{1}{x}\,dx$	$v = x$

$$\int \ln x\,dx = x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + C$$

For the definite integral:

$$\int_1^e \ln x\,dx = \Big[x\ln x - x\Big]_1^e = (e \cdot 1 - e) - (1 \cdot 0 - 1) = 1$$

The region under $y = \ln x$ from $x = 1$ to $x = e$ has area exactly $1$.

Example 3: Repeated parts (and the tabular method)

Find $\displaystyle\int x^2 \sin x\,dx$.

Solution. Take $u = x^2$ and $dv = \sin x\,dx$, so $du = 2x\,dx$ and $v = -\cos x$:

$$\int x^2 \sin x\,dx = -x^2\cos x + \int 2x\cos x\,dx$$

The new integral still needs parts. Take $u = 2x$, $dv = \cos x\,dx$, so $du = 2\,dx$, $v = \sin x$:

$$\int 2x\cos x\,dx = 2x\sin x - \int 2\sin x\,dx = 2x\sin x + 2\cos x$$

Putting it together:

$$\int x^2 \sin x\,dx = -x^2\cos x + 2x\sin x + 2\cos x + C$$

Tabular version. Differentiate $x^2$ down to $0$; integrate $\sin x$ the same number of times:

Sign	Differentiate $u$	Integrate $dv$
$+$	$x^2$	$\sin x$
$-$	$2x$	$-\cos x$
$+$	$2$	$-\sin x$
	$0$	$\cos x$

Multiply each entry in the left column by the entry one row down in the right column, using the signs: $(+)(x^2)(-\cos x) + (-)(2x)(-\sin x) + (+)(2)(\cos x)$, which gives the same answer.

Example 4: Solving for the integral

Find $\displaystyle\int e^x\cos x\,dx$.

Solution. Call the integral $I$. Take $u = \cos x$, $dv = e^x\,dx$, so $du = -\sin x\,dx$, $v = e^x$:

$$I = e^x\cos x + \int e^x\sin x\,dx$$

Do parts again on the new integral, keeping the same kind of choice ($u$ = the trig function): $u = \sin x$, $dv = e^x\,dx$, so $du = \cos x\,dx$, $v = e^x$:

$$\int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx = e^x\sin x - I$$

Substitute back and solve for $I$:

$$\begin{aligned} I &= e^x\cos x + e^x\sin x - I \\ 2I &= e^x(\sin x + \cos x) \\ I &= \tfrac{1}{2}e^x(\sin x + \cos x) + C \end{aligned}$$

Check: $\dfrac{d}{dx}\left[\tfrac{1}{2}e^x(\sin x + \cos x)\right] = \tfrac{1}{2}e^x(\sin x + \cos x) + \tfrac{1}{2}e^x(\cos x - \sin x) = e^x\cos x$. ✓

Common mistakes

Choosing u and dv backwards. With $u = e^{2x}$ and $dv = x\,dx$ in Example 1, the new integral is $\int \tfrac{1}{2}x^2 \cdot 2e^{2x}\,dx$, which is worse. If the power of $x$ goes up, switch your choice.

Sign errors in the minus sign. The formula has $uv - \int v\,du$. When $v$ is itself negative (like $v = -\cos x$), the subtraction becomes addition. Put brackets around $\int v\,du$ before simplifying.

Switching choices on the second round. In Example 4, if you used $u = \cos x$ the first time but $u = e^x$ the second time, you would just undo your first step and get $I = I$. Keep the same type of function as $u$ each time.

Forgetting to evaluate uv at the limits. In a definite integral, the $uv$ term needs $\Big[uv\Big]_a^b$ too, not just the leftover integral.

Dropping the + C or adding it too early. Leave the constant off $v$; put a single $+ C$ on the final indefinite answer.

Forcing parts when substitution works. $\int x e^{x^2}\,dx$ is a $u$-substitution ($u = x^2$), not parts. Always check for an inside function and its derivative first.

Practice

1. (Warm-up) Find $\displaystyle\int x\cos x\,dx$.

Solution

$u = x$, $dv = \cos x\,dx$, so $du = dx$, $v = \sin x$:

$$\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x + C$$

2. (Warm-up) Find $\displaystyle\int x e^{-x}\,dx$.

Solution

$u = x$, $dv = e^{-x}\,dx$, so $du = dx$, $v = -e^{-x}$:

$$\int x e^{-x}\,dx = -x e^{-x} - \int \left(-e^{-x}\right)dx = -x e^{-x} - e^{-x} + C$$

3. (Warm-up) Find $\displaystyle\int x\ln x\,dx$.

Solution

Logarithm first in LIATE: $u = \ln x$, $dv = x\,dx$, so $du = \tfrac{1}{x}\,dx$, $v = \tfrac{1}{2}x^2$:

$$\int x\ln x\,dx = \tfrac{1}{2}x^2\ln x - \int \tfrac{1}{2}x^2 \cdot \tfrac{1}{x}\,dx = \tfrac{1}{2}x^2\ln x - \tfrac{1}{4}x^2 + C$$

4. (Core) Evaluate $\displaystyle\int_0^1 x e^x\,dx$.

Solution

$u = x$, $dv = e^x\,dx$, so $du = dx$, $v = e^x$:

$$\int_0^1 x e^x\,dx = \Big[x e^x\Big]_0^1 - \int_0^1 e^x\,dx = e - \Big[e^x\Big]_0^1 = e - (e - 1) = 1$$

5. (Core) Find $\displaystyle\int x^2 e^x\,dx$.

Solution

Tabular method with $u = x^2$, $dv = e^x\,dx$:

Sign	Differentiate	Integrate
$+$	$x^2$	$e^x$
$-$	$2x$	$e^x$
$+$	$2$	$e^x$
	$0$	$e^x$

$$\int x^2 e^x\,dx = x^2 e^x - 2x e^x + 2e^x + C = e^x\left(x^2 - 2x + 2\right) + C$$

6. (Core) Find $\displaystyle\int \arctan x\,dx$.

Solution

As with $\ln x$, use $u = \arctan x$, $dv = dx$, so $du = \dfrac{1}{1 + x^2}\,dx$, $v = x$:

$$\int \arctan x\,dx = x\arctan x - \int \frac{x}{1 + x^2}\,dx$$

The last integral is a substitution ($w = 1 + x^2$, $x\,dx = \tfrac{1}{2}\,dw$):

$$\int \arctan x\,dx = x\arctan x - \tfrac{1}{2}\ln\left(1 + x^2\right) + C$$

7. (Core) A cyclist’s velocity is $v(t) = t e^{-t/2}$ metres per second, for $t$ in seconds. Find the distance she travels from $t = 0$ to $t = 4$. Give an exact answer and a decimal to 3 places.

Solution

Since $v(t) \ge 0$, distance $= \displaystyle\int_0^4 t e^{-t/2}\,dt$. Take $u = t$, $dv = e^{-t/2}\,dt$, so $du = dt$, $v = -2e^{-t/2}$:

$$\begin{aligned} \int_0^4 t e^{-t/2}\,dt &= \Big[-2t e^{-t/2}\Big]_0^4 + \int_0^4 2e^{-t/2}\,dt \\ &= -8e^{-2} + \Big[-4e^{-t/2}\Big]_0^4 \\ &= -8e^{-2} - 4e^{-2} + 4 = 4 - 12e^{-2} \approx 2.376 \text{ m} \end{aligned}$$

8. (Challenge) Find $\displaystyle\int e^{2x}\sin x\,dx$.

Solution

Let $I$ be the integral. Use $u = \sin x$, $dv = e^{2x}\,dx$, so $du = \cos x\,dx$, $v = \tfrac{1}{2}e^{2x}$:

$$I = \tfrac{1}{2}e^{2x}\sin x - \tfrac{1}{2}\int e^{2x}\cos x\,dx$$

Again with the trig function as $u$: $u = \cos x$, $dv = e^{2x}\,dx$, so $du = -\sin x\,dx$, $v = \tfrac{1}{2}e^{2x}$:

$$\int e^{2x}\cos x\,dx = \tfrac{1}{2}e^{2x}\cos x + \tfrac{1}{2}I$$

So

$$\begin{aligned} I &= \tfrac{1}{2}e^{2x}\sin x - \tfrac{1}{4}e^{2x}\cos x - \tfrac{1}{4}I \\ \tfrac{5}{4}I &= \tfrac{1}{4}e^{2x}(2\sin x - \cos x) \\ I &= \tfrac{1}{5}e^{2x}(2\sin x - \cos x) + C \end{aligned}$$

9. (Challenge) Find $\displaystyle\int (\ln x)^2\,dx$.

Solution

$u = (\ln x)^2$, $dv = dx$, so $du = \dfrac{2\ln x}{x}\,dx$, $v = x$:

$$\int (\ln x)^2\,dx = x(\ln x)^2 - \int 2\ln x\,dx$$

From Example 2, $\int \ln x\,dx = x\ln x - x$, so

$$\int (\ln x)^2\,dx = x(\ln x)^2 - 2x\ln x + 2x + C$$