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What is Calculus II

Integration
by Parts?

The product rule reversed — splitting an integral into two parts to handle products of functions that u-substitution can't touch.

Simple Definition

When you need to integrate a product of two functions — like \(x\,e^x\) or \(x\,\ln x\) — neither basic rules nor u-substitution will work directly. Integration by Parts (IBP) is the technique for these cases.

The formula is: \(\int u\,dv = uv - \int v\,du\). You split the integrand into two pieces: \(u\) (which you differentiate) and \(dv\) (which you integrate). The result trades your hard integral for (hopefully) an easier one on the right.

The secret is in the choice. Pick \(u\) to be the function that gets simpler when differentiated, and \(dv\) to be the rest — something you can actually integrate.

The Formula

Integration by Parts

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

Derived from the product rule: \(\frac{d}{dx}[uv] = u'v + uv'\). Rearranging and integrating both sides gives the formula. The \(uv\) term on the right requires no integration — just evaluate. The remaining integral \(\int v\,du\) should be simpler than what you started with.

Intuition

🔄

The product rule, integrated

The product rule says \((uv)' = u'v + uv'\). Integrate both sides: \(uv = \int u'v\,dx + \int uv'\,dx\). Rearrange: \(\int uv'\,dx = uv - \int u'v\,dx\). In \(dv\) notation, \(v' dx = dv\) and \(u' dx = du\), giving \(\int u\,dv = uv - \int v\,du\). You're literally un-doing the product rule.

🔁

When to apply it twice

Sometimes \(\int v\,du\) is still hard, and you apply IBP again. For \(\int x^2 e^x\,dx\), you need it twice. For \(\int e^x \sin x\,dx\), you apply it twice and the original integral reappears — then you solve algebraically for it. This "circular" trick is beautiful and exam-worthy.

Choosing u — The LIATE Rule

LIATE — priority order for choosing u

L

Logarithms

\(\ln x,\ \log x\)

u = this

I

Inverse trig

\(\arctan x,\ \arcsin x\)

u = this

A

Algebraic

\(x^n,\ x^2+1\)

u = this

T

Trigonometric

\(\sin x,\ \cos x\)

dv = this

E

Exponential

\(e^x,\ 2^x\)

dv = this

Step by Step — \(\int x\,e^x\,dx\)

	Choose	Compute
u	\(u = x\)	\(du = dx\)
dv	\(dv = e^x\,dx\)	\(v = e^x\)

→

Apply the formula

\(\displaystyle\int x\,e^x\,dx = uv - \int v\,du = x\,e^x - \int e^x\,dx\)

→

Solve remaining integral

\(= x\,e^x - e^x + C = e^x(x-1) + C\) ✓

Quick Examples

Example 1

\(\displaystyle\int x\cos x\,dx\)

\(u=x,\ dv=\cos x\,dx\)
\(v=\sin x\)
\(= x\sin x - \displaystyle\int \sin x\,dx\)
\(= x\sin x + \cos x + C\)

Example 2

\(\displaystyle\int \ln x\,dx\)

\(u=\ln x,\ dv=dx\)
\(du=\frac{1}{x}dx,\ v=x\)
\(= x\ln x - \displaystyle\int dx\)
\(= x\ln x - x + C\)

Example 3 — Apply twice

\(\displaystyle\int x^2 e^x\,dx\)

IBP once: \(x^2 e^x - 2\displaystyle\int xe^x\,dx\)
IBP again on \(\int xe^x\,dx\):
\(= e^x(x^2-2x+2)+C\)

Example 4 — Circular IBP

\(\displaystyle\int e^x\sin x\,dx\)

IBP twice → \(I = e^x\sin x - e^x\cos x - I\)
Solve: \(2I = e^x(\sin x - \cos x)\)
\(I = \frac{e^x(\sin x - \cos x)}{2}+C\)

Common Mistakes

✗ Choosing u and dv so that the new integral is harder

If \(\int v\,du\) is harder than what you started with, swap your choices. For \(\int x\ln x\,dx\): choose \(u = \ln x\) (simplifies to \(\frac{1}{x}\)) and \(dv = x\,dx\) (integrates to \(\frac{x^2}{2}\)). Choosing the opposite makes the new integral impossible.

✗ Forgetting the minus sign

\(\int u\,dv = uv \mathbf{-} \int v\,du\). The minus sign is essential and easy to drop when writing out a long calculation. Missing it negates your entire answer.

✗ Choosing dv as something you can't integrate

You must be able to find \(v\) from \(dv\). If you let \(dv = \ln x\,dx\), finding \(v\) requires integrating \(\ln x\) — which itself needs integration by parts! Choose \(dv\) as something with a known antiderivative: \(e^x\), \(\sin x\), \(x^n\), etc.

Related Concepts

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