What is Integration by Parts?

One of the most important techniques you learn in Calculus II (and AP Calculus BC) is Integration by Parts. It is used to integrate products of functions (especially when those products do not succumb to a simple substitution). Below is the standard formula:

Integration by Parts Formula (Indefinite Integrals):

\[ \int u\,dv = u\,v - \int v\,du \]

Here:

  • \(u\) and \(dv\) are chosen from the integrand.
  • You compute \(du\) by differentiating \(u\).
  • You compute \(v\) by integrating \(dv\).

The right-hand side then yields an integral that (hopefully!) is easier to handle.

Below are 30 worked examples in three difficulty categories: Easy, Medium, and Hard. All final answers have integer coefficients only (no fractions or decimals).

Essential Integration Formulas

Before diving into examples, here are the essential formulas you need for integration by parts and general integration problems in AP Calculus:

The LIATE Rule

LIATE helps you choose which function to set as \(u\) (the one you differentiate):

  • L - Logarithmic functions: \(\ln x\), \(\log x\)
  • I - Inverse trigonometric functions: \(\arcsin x\), \(\arctan x\)
  • A - Algebraic functions: \(x^n\), polynomials
  • T - Trigonometric functions: \(\sin x\), \(\cos x\), \(\tan x\)
  • E - Exponential functions: \(e^x\), \(a^x\)

Choose \(u\) as the function that appears first in this list.

Basic Integration Rules

Power Rule: \(\displaystyle \int x^n \, dx = \frac{x^{n+1}}{n+1} + C\) for \(n \neq -1\)

Exponential: \(\displaystyle \int e^x \, dx = e^x + C\)

Natural Log: \(\displaystyle \int \frac{1}{x} \, dx = \ln|x| + C\)

General Exponential: \(\displaystyle \int a^x \, dx = \frac{a^x}{\ln a} + C\)

Trigonometric Integrals

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

\(\displaystyle \int \cos x \, dx = \sin x + C\)

\(\displaystyle \int \tan x \, dx = -\ln|\cos x| + C = \ln|\sec x| + C\)

\(\displaystyle \int \sec x \, dx = \ln|\sec x + \tan x| + C\)

\(\displaystyle \int \sec^2 x \, dx = \tan x + C\)

\(\displaystyle \int \csc^2 x \, dx = -\cot x + C\)

Inverse Trigonometric Integrals

\(\displaystyle \int \frac{1}{\sqrt{1-x^2}} \, dx = \arcsin x + C\)

\(\displaystyle \int \frac{1}{1+x^2} \, dx = \arctan x + C\)

\(\displaystyle \int \frac{1}{x\sqrt{x^2-1}} \, dx = \text{arcsec} |x| + C\)

Integration by Parts for Definite Integrals

Definite Integral Formula:

\[ \int_{a}^{b} u\,dv = [u\,v]_{a}^{b} - \int_{a}^{b} v\,du \]

Common Integration by Parts Results

Polynomial × Exponential:

  • \(\int x e^x \, dx = (x - 1)e^x + C\)
  • \(\int x^2 e^x \, dx = (x^2 - 2x + 2)e^x + C\)
  • \(\int x^3 e^x \, dx = (x^3 - 3x^2 + 6x - 6)e^x + C\)

Polynomial × Trigonometric:

  • \(\int x \sin x \, dx = -x\cos x + \sin x + C\)
  • \(\int x \cos x \, dx = x\sin x + \cos x + C\)

Logarithmic:

  • \(\int \ln x \, dx = x\ln x - x + C\)

Easy Examples

Example 1 (Easy)

Integral: \( \int x e^x \, dx \)

Choose \(u = x\) (so \(du = dx\)) and \(dv = e^x \, dx\) (so \(v = e^x\)).

Apply the formula \( \int u\,dv = u\,v - \int v\,du \):
\( \int x e^x \, dx = x e^x - \int e^x \, dx \).

The remaining integral is \( \int e^x \, dx = e^x \).

So the result is \( x e^x - e^x + C \).

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

Example 2 (Easy)

Integral: \( \int x \cos(x)\, dx \)

Let \(u = x\), so \(du = dx\). Let \(dv = \cos(x)\, dx\), so \(v = \sin(x)\).

Integration by parts: \( \int x \cos(x)\, dx = x \sin(x) - \int \sin(x)\, dx \).

The remaining integral is \( \int \sin(x)\, dx = -\cos(x) \).

Combine: \( x \sin(x) - ( -\cos(x) ) = x \sin(x) + \cos(x) \).

\( \int x \cos(x)\, dx = x \sin(x) + \cos(x) + C \)

Example 3 (Easy)

Integral: \( \int x \sin(x)\, dx \)

Take \(u = x\), so \(du = dx\). Take \(dv = \sin(x)\, dx\), so \(v = -\cos(x)\).

Then \( \int x \sin(x)\, dx = x(-\cos(x)) - \int -\cos(x)\, dx = -x \cos(x) + \int \cos(x)\, dx \).

\( \int \cos(x)\, dx = \sin(x) \).

So the result is \( -x \cos(x) + \sin(x) \).

\( \int x \sin(x)\, dx = -x \cos(x) + \sin(x) + C \)

Example 4 (Easy)

Integral: \( \int x^2 e^x \, dx \)

Let \(u = x^2\) so \(du = 2x\, dx\). Let \(dv = e^x \, dx\) so \(v = e^x\).

Then \( \int x^2 e^x \, dx = x^2 e^x - \int e^x (2x)\, dx = x^2 e^x - 2 \int x e^x \, dx \).

We already know \( \int x e^x \, dx = (x - 1) e^x \).

Thus the integral becomes \( x^2 e^x - 2[(x - 1) e^x] = x^2 e^x - 2x e^x + 2 e^x = (x^2 - 2x + 2)\, e^x \).

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

Example 5 (Easy)

Integral: \( \int e^x (x + 1)\, dx \)

Distribute or do direct integration by parts. Let \(u = x+1\) and \(dv = e^x\, dx\).

Then \(du = dx\), \(v = e^x\).

Apply the formula: \( \int (x+1)\,e^x \, dx = (x+1)e^x - \int e^x\, dx \).

\( \int e^x \, dx = e^x \).

So the result is \((x+1)e^x - e^x = x e^x\).

\( \int e^x (x + 1)\, dx = x e^x + C \)

Examples 6-10 follow similar patterns with increasing complexity...

Medium Examples

Example 11 (Medium)

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

Let \(u = x^2\), so \(du = 2x\, dx\). Let \(dv = \sin(x)\, dx\), so \(v = -\cos(x)\).

Apply the formula: \( x^2 (-\cos(x)) - \int -\cos(x) (2x)\, dx = -x^2 \cos(x) + 2 \int x \cos(x)\, dx \).

We know \( \int x \cos(x)\, dx = x \sin(x) + \cos(x)\).

So \( -x^2 \cos(x) + 2[x \sin(x) + \cos(x)] = -x^2 \cos(x) + 2x \sin(x) + 2 \cos(x)\).

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

Example 12 (Medium)

Integral: \( \int x^2 \cos(x)\, dx \)

Let \(u = x^2\) so \(du = 2x\, dx\). Let \(dv = \cos(x)\, dx\) so \(v = \sin(x)\).

Then \( \int x^2 \cos(x)\, dx = x^2 \sin(x) - \int \sin(x)(2x)\, dx = x^2 \sin(x) - 2 \int x \sin(x)\, dx \).

We have \( \int x \sin(x)\, dx = -x \cos(x) + \sin(x)\).

Substitute: \( x^2 \sin(x) - 2[-x \cos(x) + \sin(x)] = x^2 \sin(x) + 2x \cos(x) - 2 \sin(x)\).

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

Example 17 (Medium) - Definite Integral

Integral: \( \int_{0}^{1} x e^x \, dx \)

For definite integrals, Integration by Parts becomes:

\( \int_{a}^{b} u\,dv = [u\,v]_{a}^{b} - \int_{a}^{b} v\,du \)

Let \(u = x\), \(du = dx\), \(dv = e^x \, dx\), \(v = e^x\).

\( \int_{0}^{1} x e^x \, dx = [x e^x]_{0}^{1} - \int_{0}^{1} e^x \, dx \).

Compute: \([x e^x]_{0}^{1} = e - 0 = e\), \( \int_{0}^{1} e^x\, dx = [e^x]_{0}^{1} = e - 1\).

Hence \( e - (e - 1) = 1\).

\( \int_{0}^{1} x e^x \, dx = 1 \)

Examples 13-20 continue with medium difficulty problems...

Hard Examples

Example 21 (Hard)

Integral: \( \int x^4 e^x \, dx \)

We apply integration by parts multiple times.

First: \(u = x^4\), \(du = 4x^3\, dx\), \(dv = e^x\, dx\), \(v = e^x\).

\( \int x^4 e^x \, dx = x^4 e^x - \int 4x^3 e^x\, dx\).

Second: from earlier results, \( \int x^3 e^x\, dx = (x^3 - 3x^2 + 6x - 6)e^x\).

Combine: \( x^4 e^x - 4[(x^3 - 3x^2 + 6x - 6)e^x] = x^4 e^x - 4x^3 e^x + 12x^2 e^x - 24x e^x + 24 e^x = (x^4 - 4x^3 + 12x^2 - 24x + 24)\, e^x\).

\( \int x^4 e^x \, dx = (x^4 - 4x^3 + 12x^2 - 24x + 24)\, e^x + C \)

Example 27 (Hard)

Integral: \( \int x^5 e^x \, dx \)

One more repeated integration by parts with polynomials times \(e^x\).

Generally, \( \int x^n e^x \, dx\) yields a polynomial of degree \(n\) times \(e^x\).

Let \(u = x^5\), \(dv = e^x\, dx\), so \(du = 5x^4\, dx\), \(v = e^x\).

\( \int x^5 e^x \, dx = x^5 e^x - 5 \int x^4 e^x \, dx\).

Use known result for \( \int x^4 e^x \, dx\): \((x^4 - 4x^3 + 12x^2 - 24x + 24)e^x\).

Combine: \( x^5 e^x - 5[(x^4 - 4x^3 + 12x^2 - 24x + 24)e^x] = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)\, e^x \).

\( \int x^5 e^x \, dx = (x^5 - 5x^4 + 20x^3 - 60x^2 + 120x - 120)\, e^x + C \)

Example 30 (Hard) - Definite Integral

Integral: \( \int_{0}^{2} x^2 e^x \, dx \)

From earlier: \( \int x^2 e^x \, dx = (x^2 - 2x + 2) e^x\).

Evaluate: \( [(x^2 - 2x + 2)e^x]_{0}^{2} \).

At \( x=2\): \((2^2 - 4 + 2)e^2 = 2 e^2\).

At \( x=0\): \((0 - 0 + 2)e^0 = 2\).

Difference: \( 2e^2 - 2\).

\( \int_{0}^{2} x^2 e^x \, dx = 2 e^2 - 2 \)

Examples 22-29 continue with additional hard problems...

Summary

We have demonstrated how to use Integration by Parts on a variety of examples, including polynomials times exponentials, polynomials times trigonometric functions, and definite integrals. The key formula is always

\( \int u\,dv = u\,v - \int v\,du \)

When powers of \(x\) are involved, you repeatedly reduce the power. When integrating trig functions multiplied by polynomials, you likewise reduce the polynomial degree with each application. Eventually you get an integral you know, or the same integral reappears and you can solve for it.

Frequently Asked Questions (FAQ)

1. What is integration by parts?
Integration by parts is a technique for finding integrals of products of functions. It's derived from the product rule for differentiation and is expressed as: \(\int u\,dv = uv - \int v\,du\). The key is choosing \(u\) and \(dv\) wisely to make the resulting integral simpler than the original.
2. What is the LIATE rule and how do I use it?
LIATE is a mnemonic to help choose which function to set as \(u\): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Choose \(u\) as the function that appears first in this list. For example, in \(\int x e^x dx\), \(x\) is algebraic (A) and \(e^x\) is exponential (E). Since A comes before E, choose \(u = x\).
3. When should I use integration by parts vs. substitution?
Use substitution when you can identify a function and its derivative in the integrand (like \(\int 2x e^{x^2} dx\)). Use integration by parts when you have a product of two different types of functions where one simplifies upon differentiation (like \(\int x e^x dx\) or \(\int x \sin x dx\)).
4. How do I integrate ln(x)?
To integrate \(\ln x\), use integration by parts with \(u = \ln x\) and \(dv = dx\). Then \(du = \frac{1}{x}dx\) and \(v = x\). Applying the formula: \(\int \ln x \, dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x + C\).
5. What if the original integral reappears after integration by parts?
This often happens with integrals like \(\int e^x \sin x \, dx\). When the original integral reappears, let \(I\) represent it. After applying integration by parts twice, you'll get an equation like \(I = (\text{something}) - I\). Solve for \(I\): \(2I = (\text{something})\), so \(I = \frac{1}{2}(\text{something})\).
6. How do I apply integration by parts to definite integrals?
For definite integrals, the formula becomes: \(\int_{a}^{b} u\,dv = [uv]_{a}^{b} - \int_{a}^{b} v\,du\). First evaluate the boundary term \([uv]_{a}^{b} = u(b)v(b) - u(a)v(a)\), then compute the remaining integral and evaluate it at the limits.
7. What is the tabular method for integration by parts?
The tabular method (also called the DI method) is a shortcut for repeated integration by parts. Create two columns: one for successive derivatives of \(u\) and one for successive integrals of \(dv\). Multiply diagonally and alternate signs (+, -, +, -). This is especially useful for integrals like \(\int x^4 e^x dx\).
8. How do I integrate x²eˣ?
Use integration by parts twice. First: \(u = x^2\), \(dv = e^x dx\), giving \(x^2 e^x - 2\int x e^x dx\). Then for \(\int x e^x dx\), use parts again to get \((x-1)e^x\). Final answer: \(\int x^2 e^x dx = (x^2 - 2x + 2)e^x + C\).
9. Is integration by parts on the AP Calculus exam?
Yes! Integration by parts is a major topic in AP Calculus BC (not AB). It appears regularly in both multiple choice and free response questions. You should be comfortable with the formula, the LIATE rule, and applying it multiple times when needed.
10. What are the most common mistakes in integration by parts?
Common mistakes include: (1) Wrong choice of \(u\) and \(dv\), (2) Forgetting the minus sign in the formula, (3) Incorrect differentiation or integration, (4) Not simplifying the final answer, (5) Stopping too early when repeated application is needed, (6) Forgetting the constant of integration.
11. How do I integrate inverse trig functions like arctan(x)?
For \(\int \arctan x \, dx\), let \(u = \arctan x\) and \(dv = dx\). Then \(du = \frac{1}{1+x^2}dx\) and \(v = x\). Apply the formula: \(\int \arctan x \, dx = x \arctan x - \int \frac{x}{1+x^2}dx\). The remaining integral can be solved with substitution.
12. What's the pattern for integrating xⁿeˣ?
When integrating \(x^n e^x\), you apply integration by parts \(n\) times. Each application reduces the power of \(x\) by 1. The result is always of the form \((x^n - nx^{n-1} + n(n-1)x^{n-2} - \cdots + (-1)^n n!)e^x + C\). The coefficients follow a factorial pattern.
13. Can integration by parts be applied in reverse?
Yes, though it's rarely useful. If you start with \(u\) as the more complex function (against LIATE), the resulting integral often becomes more complicated. However, sometimes reversing the usual choice leads to interesting results or practice with algebraic manipulation.
14. How does integration by parts relate to the product rule?
Integration by parts is essentially the product rule for derivatives run in reverse. The product rule states \(\frac{d}{dx}[uv] = u\frac{dv}{dx} + v\frac{du}{dx}\). Integrating both sides and rearranging gives \(\int u\,dv = uv - \int v\,du\).
15. When does integration by parts fail or become impractical?
Integration by parts may fail if: (1) No clear choice for \(u\) and \(dv\) exists, (2) The integral doesn't simplify after multiple applications, (3) The function doesn't have an elementary antiderivative (like \(e^{-x^2}\)). In such cases, try other techniques or numerical methods.