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\(\int\)
\(uv\)

Integration
By Parts

Master the technique of turning complex integrals into simpler ones by strategically splitting the integrand into two factors.

AP Calculus BC

Why Do We Need Integration By Parts?

Many integrals involve products of functions that don't yield to basic techniques like substitution. For example, how would you evaluate \(\int x \, e^x \, dx\)?

There is no product rule for integration. But we can reverse the product rule for derivatives to create a powerful method called Integration By Parts.

From the Product Rule to IBP

Recall the product rule for derivatives:

Now integrate both sides with respect to \(x\):

Rearranging gives us the Integration By Parts formula. This algebraic rearrangement is the entire derivation: nothing more, nothing less.

The IBP Formula

\(u\)
A function you choose to differentiate
\(dv\)
The remaining piece you integrate
\(v\,du\)
The new integral: ideally simpler

The key insight: you trade one integral (\(\int u\,dv\)) for another (\(\int v\,du\)). The method succeeds when the new integral is easier than the original.

Choosing \(u\): The LIATE Rule

The hardest part of IBP is choosing which factor to call \(u\). The LIATE mnemonic ranks function types: pick \(u\) from higher on the list.

L
Logarithmic: \(\ln x,\; \log x\)
I
Inverse trig: \(\arctan x,\; \arcsin x\)
A
Algebraic: \(x^n,\; x^2 + 3x\)
T
Trigonometric: \(\sin x,\; \cos x\)
E
Exponential: \(e^x,\; 2^x\)

IBP vs U-Substitution

Integration By Parts U-Substitution
Reverses Product rule Chain rule
Best for Products of different types Compositions \(f(g(x))\,g'(x)\)
You choose \(u\) and \(dv\) \(u = g(x)\)
Goal Simpler resulting integral Eliminate inner function
Example \(\int x\,\cos x\,dx\) \(\int 2x\,\cos(x^2)\,dx\)

How to Apply IBP Step-by-Step

1
Identify the product. Confirm your integrand is a product of two distinct function types.
2
Choose \(u\) and \(dv\). Use LIATE to pick \(u\) (the part to differentiate). Everything else is \(dv\).
3
Compute \(du\) and \(v\). Differentiate \(u\) to get \(du\). Integrate \(dv\) to get \(v\).
4
Substitute into the formula: \(\int u\,dv = u\,v - \int v\,du\).
5
Evaluate the new integral. If it is still complex, apply IBP again or use another technique.

The IBP Flow

\(\int u\,dv\)
\(u\) → differentiate = \(du\)
\(dv\) → integrate = \(v\)
\(u \cdot v \;-\; \int v\,du\)
 Evaluate the new integral

Example 1: \(\int x\,e^x\,dx\)

1
Choose parts: Let \(u = x\) (algebraic, higher on LIATE) and \(dv = e^x\,dx\)
2
Differentiate and integrate: \(du = dx\), \(v = e^x\)
3
Apply the formula: \(\int x\,e^x\,dx = x\,e^x - \int e^x\,dx\)
4
Evaluate: \(= x\,e^x - e^x + C\)

Example 2: \(\int x^2\,\sin x\,dx\)

This requires applying IBP twice because the algebraic factor has degree 2.

1
First pass: \(u = x^2,\; dv = \sin x\,dx\) so \(du = 2x\,dx,\; v = -\cos x\)
\(\Rightarrow x^2(-\cos x) - \int (-\cos x)(2x)\,dx = -x^2\cos x + 2\!\int x\cos x\,dx\)
2
Second pass on \(\int x\cos x\,dx\): \(u = x,\; dv = \cos x\,dx\) so \(du = dx,\; v = \sin x\)
\(\Rightarrow x\sin x - \int \sin x\,dx = x\sin x + \cos x\)
3
Combine: \(-x^2\cos x + 2(x\sin x + \cos x) + C\)

Example 3: \(\int \ln x\,dx\)

This integral has only one factor, but we can set \(dv = dx\) to make IBP work. This is a classic trick.

1
Choose: \(u = \ln x\), \(dv = dx\)
2
Compute: \(du = \frac{1}{x}\,dx\), \(v = x\)
3
Apply: \(\displaystyle x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - \int 1\,dx\)
4
Evaluate: \(= x\ln x - x + C\)

The Cyclic (Boomerang) Case

Sometimes IBP leads back to the original integral. This isn't failure: it's a feature! You can solve for the integral algebraically.

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

The Tabular Method (Quick IBP)

When \(u\) is a polynomial, repeated IBP follows a predictable pattern. The tabular method organizes this into a table.

Example: \(\int x^3 e^{2x}\,dx\). Let \(u = x^3\) and \(dv = g(x)\,dx = e^{2x}\,dx\):

Sign Derivatives of \(u\) \(g(x)\) & its Integrals
+\(x^3\)\(e^{2x}\)
−\(3x^2\)\(\tfrac{1}{2}e^{2x}\)
+\(6x\)\(\tfrac{1}{4}e^{2x}\)
−\(6\)\(\tfrac{1}{8}e^{2x}\)
+\(0\)\(\tfrac{1}{16}e^{2x}\)

IBP with Definite Integrals

For definite integrals, apply the bounds to the \(uv\) term and the remaining integral:

Example: Evaluate \(\displaystyle\int_0^1 x\,e^x\,dx\).

1
From Example 1: antiderivative is \(e^x(x - 1)\)
2
Evaluate: \(\bigl[e^x(x-1)\bigr]_0^1 = e^1(1-1) - e^0(0-1) = 0 - (-1) = 1\)

Common Mistakes to Avoid

✕ Wrong \(u\) choice
Choosing \(u = e^x\) and \(dv = x\,dx\) for \(\int x e^x\,dx\) makes the integral harder, not easier.
✕ Forgetting the minus sign
The formula is \(uv \mathbf{-} \int v\,du\). Dropping the negative flips the entire answer.
✕ Omitting \(+C\)
For indefinite integrals, the constant of integration is required. Missing it loses points on the AP exam.
✕ Inconsistent cycling
In the cyclic case, switching which factor is \(u\) mid-process prevents the original integral from reappearing.
✕ Ignoring simpler methods
IBP for \(\int 2x \cos(x^2)\,dx\) is overkill: simple u-substitution works instantly.

Summary: When and How

Integral Type Choose \(u\) Choose \(dv\) Notes
\(\int x^n e^{ax}\,dx\) \(x^n\) \(e^{ax}\,dx\) Tabular if \(n \geq 2\)
\(\int x^n \sin x\,dx\) \(x^n\) \(\sin x\,dx\) Repeat \(n\) times
\(\int \ln x\,dx\) \(\ln x\) \(dx\) Classic trick
\(\int e^x \sin x\,dx\) Either Either Cyclic: solve for \(I\)
\(\int \arctan x\,dx\) \(\arctan x\) \(dx\) Inverse trig = \(u\)

Question 1 of 5

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

Question 2 of 5

When evaluating \(\displaystyle\int x^2 e^x\,dx\) by parts, what should you choose for \(u\)?

Question 3 of 5

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

Question 4 of 5

Which integral requires the cyclic (boomerang) technique when solved by parts?

Question 5 of 5

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