3.1 Integration by Parts

Big idea

Substitution is the integral version of the chain rule. This section supplies the integral version of the product rule, and it is the tool you reach for when the integrand is a product of two unrelated pieces: a polynomial times an exponential, a power times a logarithm, an exponential times a sine.

The derivation is one line. Differentiating a product gives $(uv)' = u'v + uv'$. Integrate both sides and rearrange so that one of the two pieces sits alone on the left. What comes out is a trade: you exchange the integral you were handed for a different integral, plus a term you get for free. Nothing is evaluated yet. The method only pays off if the new integral is easier than the old one.

That condition is the whole art of the technique. Substitution either works or it does not, and you find out quickly. Integration by parts always applies, so the question is never whether you can use it but whether the split you chose moved you forward. A bad split turns $\int x e^{x}\,dx$ into something with $x^2$ in it, which is worse. A good split lowers the power of the polynomial by one and leaves the exponential alone.

Three patterns cover most of what you will meet. Sometimes one application finishes the job. Sometimes the new integral has the same shape as the old one with a smaller exponent, and you apply the rule again, and again, until the polynomial factor is gone. And sometimes the original integral reappears on the right side unchanged, which looks like failure but is actually an equation you can solve for the unknown integral.

Decoder

If $u$ and $v$ are differentiable, then the integral of $u\,dv$ equals $uv$ minus the integral of $v\,du$.

Every symbol here is a choice you make. You split the integrand into two parts: one part becomes $u$, and everything else, including the $dx$, becomes $dv$. Then you differentiate $u$ to get $du$ and antidifferentiate $dv$ to get $v$. The formula assembles those four objects.

Two constraints govern the split. You must be able to antidifferentiate $dv$, or you cannot produce $v$ at all, and differentiating $u$ should simplify it. The mnemonic LIATE ranks function types by how much they improve under differentiation: logarithmic, inverse trigonometric, algebraic, trigonometric, exponential. Whichever type appears earlier in that list is the better choice for $u$. A logarithm differentiates into a power, which is a genuine simplification; an exponential differentiates into itself, so it makes a fine $dv$ and a poor $u$.

The classic mistake is losing the minus sign in front of the second integral, and the second classic mistake is attaching a constant of integration to $v$ too early. You do not need one there. Any antiderivative of $dv$ works, so take the simplest, and save the $+ C$ for the very end.

Definitions and results

The formula. For differentiable $u$ and $v$,

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

Where it comes from. The product rule says $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$. Integrating both sides gives $uv = \int u\,dv + \int v\,du$, and moving one integral across is the formula. No approximation is involved; it is the product rule read backwards.

The definite version. On an interval $[a,b]$ the boundary term is evaluated and the integral limits carry over:

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

Evaluate the bracket before you touch the remaining integral, so that a term that vanishes at both ends disappears early.

Choosing $u$ by LIATE. Rank the factors as logarithmic, inverse trigonometric, algebraic, trigonometric, exponential, and let $u$ be the one from the earliest category. The remaining factor together with $dx$ is $dv$. This is a heuristic, not a theorem, but it picks correctly in nearly every standard problem.

Integrating a lone logarithm. When the integrand does not look like a product, you can still manufacture one by taking $dv = dx$. With $u = \ln x$ and $v = x$,

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

The same trick handles $\arctan x$ and $\arcsin x$.

Repeated application. If $u$ is a polynomial of degree $n$ and $dv$ is an exponential or a sine or cosine, each application lowers the degree by one, so $n$ applications finish the problem. Keep $u$ in the same category every time; switching mid-problem undoes the previous step and returns you to where you started.

The integral that returns. For $\int e^{ax}\sin bx\,dx$ and its relatives, two applications reproduce the original integral with a constant in front. Call the unknown integral $I$, collect the copies of $I$ on one side, and divide. The answer comes from algebra, not from a third application.

Substitution first. Parts and substitution are not rivals. An integral such as $\int x^3 e^{x^2}\,dx$ yields to a substitution that turns it into a parts problem. Always check whether a substitution cleans up the integrand before committing to a split.

Worked examples

A polynomial times an exponential

Evaluate $\int x e^{3x}\,dx$.

By LIATE the algebraic factor outranks the exponential, so take $u = x$ and $dv = e^{3x}\,dx$. Then $du = dx$ and $v = \frac{1}{3}e^{3x}$:

$$ \int x e^{3x}\,dx = \frac{x}{3}e^{3x} - \frac{1}{3}\int e^{3x}\,dx = \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C $$

Check by differentiating. The first term gives $\frac{1}{3}e^{3x} + x e^{3x}$ and the second gives $-\frac{1}{3}e^{3x}$. The two constant-coefficient terms cancel and $x e^{3x}$ survives.

Applying the rule twice

Evaluate $\int x^2 \cos x\,dx$.

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

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

The new integral has the same shape with the power dropped by one, so run the rule again with $u = x$ and $dv = \sin x\,dx$, giving $v = -\cos x$:

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

Substitute back and distribute the $-2$:

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

Differentiating returns $2x\sin x + x^2\cos x + 2\cos x - 2x\sin x - 2\cos x = x^2\cos x$.

Solving for the integral

Evaluate $I = \int e^{x}\sin x\,dx$.

Neither factor simplifies under differentiation, so no choice makes the new integral easier. Take $u = \sin x$ and $dv = e^x\,dx$:

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

Apply parts to the remaining integral, keeping the trigonometric factor as $u$ so the process does not reverse itself. With $u = \cos x$ and $dv = e^x\,dx$,

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

Putting the two lines together gives $I = e^x\sin x - e^x\cos x - I$, so $2I = e^x(\sin x - \cos x)$ and

$$ \int e^x\sin x\,dx = \frac{e^x}{2}\big(\sin x - \cos x\big) + C $$

Differentiating the answer gives $\frac{1}{2}e^x(\sin x - \cos x) + \frac{1}{2}e^x(\cos x + \sin x) = e^x \sin x$.

A definite integral with a logarithm

Evaluate $\int_1^2 x\ln x\,dx$.

The logarithm is the better $u$, so $u = \ln x$ and $dv = x\,dx$, giving $du = \frac{dx}{x}$ and $v = \frac{x^2}{2}$:

$$ \int_1^2 x\ln x\,dx = \left[\frac{x^2}{2}\ln x\right]_1^2 - \int_1^2 \frac{x^2}{2}\cdot\frac{1}{x}\,dx $$

The bracket is $2\ln 2 - 0$ because $\ln 1 = 0$. The remaining integral is $\int_1^2 \frac{x}{2}\,dx = \left[\frac{x^2}{4}\right]_1^2 = 1 - \frac14 = \frac34$. So the value is

$$ 2\ln 2 - \frac{3}{4} \approx 0.636 $$

A sanity check on the size: the integrand runs from $0$ at $x=1$ to $2\ln 2 \approx 1.386$ at $x=2$, so a value a little above half the maximum is reasonable.

Practice

These drills ask you to pick the split, carry out the formula, and recognize when a second pass or an algebraic solve is needed. Work the easy splits first until the LIATE ordering is automatic, then take on the ones that return to themselves.

Quiz

Five items on choosing $u$ and $dv$, repeated applications, definite integrals, and integrals that must be solved for algebraically.