MISSION · Calculus
Integration by Parts
The product rule, run backwards, with a choice to make
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The idea in one sentence
Differentiate one factor and integrate the other, trading the integral you can’t do for one you can.
\[\int u \, dv = uv - \int v \, du\]Where it comes from
The product rule says
\[\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\]Integrate both sides. The left becomes \(uv\), and rearranging gives the formula above. Integration by parts is the product rule, read right to left — just as substitution is the chain rule read backwards.
A worked example
\[\int x e^x dx\]Choose \(u = x\) and \(dv = e^x dx\). Then \(du = dx\) and \(v = e^x\):
\[\int x e^x dx = x e^x - \int e^x dx = xe^x - e^x + C\]The new integral was easier than the old one, which is the entire criterion for success.
Try the other choice to see what goes wrong. With \(u = e^x\) and \(dv = x\,dx\) you get \(\frac{x^2}{2}e^x - \int \frac{x^2}{2}e^x dx\) — a worse integral than you started with. The method still applied correctly; the choice was just bad.
Choosing u: LIATE
Pick \(u\) as whichever factor comes first in this list:
Logarithmic · Inverse trig · Algebraic · Trigonometric · Exponential
The logic is that things near the front get much simpler when differentiated, and things near the back stay manageable when integrated. \(\ln x\) becomes \(1/x\); \(e^x\) stays \(e^x\) whichever you do to it.
It’s a rule of thumb with exceptions, not a theorem, but it’s right often enough to stop you guessing.
The trick for ln x
\[\int \ln x \, dx\]There’s no product here — until you invent one. Take \(u = \ln x\) and \(dv = dx\):
\[\int \ln x \, dx = x\ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x + C\]This is the standard way to integrate a lone logarithm or a lone inverse trig function.
Going round twice
\(\int x^2 \sin x \, dx\) needs parts applied twice — the \(x^2\) drops to \(2x\), then to
- Each pass reduces the power by one. A pattern like \(\int x^n e^x dx\) gives a reduction formula that steps down to a base case.
And \(\int e^x \sin x\, dx\) does something stranger: apply parts twice and the original integral reappears on the right. Call it \(I\), and you get an equation like \(I = \text{stuff} - I\), which you solve algebraically for \(I\). It looks like a dead end and it’s actually the answer.
For definite integrals
\[\int_a^b u\,dv = \left[uv\right]_a^b - \int_a^b v\,du\]Evaluate the \(uv\) part at both limits as you go. For \(\int_0^1 xe^x dx\) that gives \([xe^x - e^x]_0^1 = (e - e) - (0 - 1) = 1\).
Common mix-ups
- Losing the minus sign before the second integral.
- Forgetting the \(+C\), or adding one to \(v\) midway.
- Choosing \(u\) to be the part that gets nastier when differentiated.
- Giving up when the original integral reappears, which is exactly when you’re one line of algebra from finishing.
- Reaching for parts when substitution would do. \(\int xe^{x^2}dx\) is a substitution; the \(x\) is the derivative of the exponent.
Check yourself
Which factor is \(u\) in \(\int xe^x dx\), and what is \(\int_0^1 xe^x dx\)?
Take u = x, since differentiating it gives 1. The antiderivative is \(xe^x - e^x\), and between 0 and 1 that gives (e − e) − (0 − 1) = 1.
LEVEL 3 · REFLECT
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