AP Calculus BC: Integration by Parts

In calculus, you can differentiate almost any function you can write down, but integration is a different story. Many integrals that look simple, like $\int x{e}^{x}dx$ or $\int \ln xdx$, resist basic rules. Integration by Parts is a powerful technique that cracks open these problems by providing a method to integrate products of functions. It is essentially the reverse engineering of the Product Rule for differentiation and is indispensable for tackling integrals involving products of algebraic, exponential, logarithmic, and trigonometric functions—common occurrences in physics and engineering contexts.

The Formula: Derived from the Product Rule

The foundation of Integration by Parts is the Product Rule for differentiation. Recall that for two differentiable functions $u(x)$ and $v(x)$, the derivative of their product is: $$\frac{d}{dx}[u(x)v(x)]=u^{\prime }(x)v(x)+u(x)v^{\prime }(x).$$ If we integrate both sides with respect to $x$, we get: $$u(x)v(x)=\int v(x)u^{\prime }(x)dx+\int u(x)v^{\prime }(x)dx.$$ Rearranging this equation gives us the classic Integration by Parts formula: $$\int udv=uv-\int vdu.$$

This formula states that the integral of $udv$ can be exchanged for the expression $uv$ minus the integral of $vdu$. The entire strategy hinges on a smart initial choice: you must identify part of the original integrand as "$u$" (which you will differentiate to find $du$) and the remaining part, including the $dx$, as "$dv$" (which you will integrate to find $v$). A poor choice can lead you to a more complicated integral than you started with.

Choosing "u" and "dv": The LIATE Rule

How do you make the correct choice for $u$ and $dv$? The LIATE mnemonic is a reliable guideline for prioritizing which function to set as $u$. LIATE stands for Logarithmic, Inverse trigonometric, Algebraic, Trigonometric, and Exponential. You should let $u$ be the function that appears first in this list.

For example, consider $\int x\cos xdx$. The integrand is a product of an Algebraic function ($x$) and a Trigonometric function ($\cos x$). According to LIATE, Algebraic (A) comes before Trigonometric (T), so we choose:

• $u=x$ (Algebraic)
• $dv=\cos xdx$ (Trigonometric)

We then differentiate $u$ to find $du=dx$ and integrate $dv$ to find $v=\sin x$. Applying the formula: $$\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+C.$$ The new integral $\int \sin xdx$ is simpler, confirming our choice was good. This systematic choice is what makes the technique manageable.

Solving Common Integral Types: Logs and Inverse Trig Functions

Integration by Parts is the primary tool for integrating logarithmic and inverse trigonometric functions, which don't have elementary antiderivatives from basic rules. For these, you typically set the entire log or inverse trig function as $u$.

Example: Integrating a Logarithm Find $\int \ln xdx$. Here, we have only one function. We treat it as $\int (\ln x)\cdot (1)dx$.

• Let $u=\ln x$ (Logarithmic, highest LIATE priority).
• Let $dv=1dx$.

Then, $du=\frac{1}{x}dx$ and $v=x$. Apply the formula: $$\int \ln xdx=x\ln x-\int x\cdot \frac{1}{x}dx=x\ln x-\int 1dx=x\ln x-x+C.$$

Example: Integrating an Inverse Tangent Find $\int \arctan xdx$. We set:

• $u=\arctan x$ (Inverse trigonometric).
• $dv=1dx$.

Then, $du=\frac{1}{1+x^2}dx$ and $v=x$. Applying the formula: $$\int \arctan xdx=x\arctan x-\int \frac{x}{1+x^2}dx.$$ The remaining integral is solved with a simple $w$-substitution ($w=1+x^2$), yielding $\frac{1}{2}\ln |1+x^2|$. Thus, the final answer is $x\arctan x-\frac{1}{2}\ln (1+x^2)+C$.

Advanced Application: Repeated Use and Tabular Integration

Some integrals require applying Integration by Parts more than once. A classic example is $\int x^2{e}^{x}dx$, where the algebraic part ($x^2$) is a polynomial. Each application reduces the power of $x$ by one. After two applications, the algebraic part becomes a constant, and the problem is solved.

For integrals involving the product of a polynomial and an exponential or sine/cosine function, tabular integration (also called the "tic-tac-toe" method) provides an efficient, organized shortcut. Here’s the process for $\int x^3{e}^{2x}dx$:

1. Create two columns. In the left column, repeatedly differentiate the polynomial ($x^3$) until you reach zero. In the right column, repeatedly integrate the other factor ($e^{2x}$).
2. Draw diagonal arrows from the left column to the right column one row down, multiplying the connected terms.
3. Alternate the signs (+, -, +, -...) on these products and sum them.

The tabular setup looks like this:

The solution is the sum of the signed products: $$\int x^3{e}^{2x}dx=+\left(x^3\cdot \frac{1}{2}{e}^{2x}\right)-\left(3x^2\cdot \frac{1}{4}{e}^{2x}\right)+\left(6x\cdot \frac{1}{8}{e}^{2x}\right)-\left(6\cdot \frac{1}{16}{e}^{2x}\right)+C.$$ Simplify to: $\frac{1}{2}{e}^{2x}\left(x^3-\frac{3}{2}{x}^2+\frac{3}{2}x-\frac{3}{4}\right)+C$. Tabular integration streamlines what would otherwise be several pages of repetitive work.

Common Pitfalls

1. Misapplying the LIATE Rule: The most frequent error is an unstrategic choice of $u$ and $dv$. For instance, in $\int x{e}^{x}dx$, choosing $u=e^x$ and $dv=xdx$ leads to $v=\frac{1}{2}{x}^2$, making the new integral $\int \frac{1}{2}{x}^2{e}^{x}dx$, which is more complex. Always consult LIATE: Algebraic ($x$) comes before Exponential ($e^x$), so $u=x$ is correct.

1. Dropping the Minus Sign or the Constant of Integration: The formula is $uv-\int vdu$. Forgetting the minus sign is a catastrophic algebraic error. Also, remember to add the constant of integration "$+C$" at the end of the process when you write the final antiderivative, not after evaluating the $uv$ term.

1. Mishandling Constant Multiples in Tabular Integration: When using the tabular method, be meticulous with coefficients from differentiation and integration. For $\int x^2{e}^{3x}dx$, integrating $e^{3x}$ gives $\frac{1}{3}{e}^{3x}$, then $\frac{1}{9}{e}^{3x}$, etc. Each coefficient must be correctly included in the product.

1. Not Recognizing When to Stop or Loop: In integrals like $\int e^x\cos xdx$, applying parts once and then again will eventually produce the original integral. The correct move is to treat the reappearing integral like a variable in an algebraic equation and solve for it, rather than continuing indefinitely.

Summary

• Integration by Parts, defined by $\int udv=uv-\int vdu$, is the integral analogue of the Product Rule and is essential for integrating products of functions.
• The LIATE rule (Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential) provides a reliable heuristic for choosing $u$, ensuring the resulting integral $\int vdu$ is simpler.
• This technique is the standard method for finding antiderivatives of single logarithmic (e.g., $\ln x$) and inverse trigonometric functions (e.g., $\arctan x$).
• For integrals requiring repeated application, particularly polynomials times exponentials or trig functions, tabular integration offers a structured, efficient shortcut to the final answer.
• Success depends on careful selection of $u$, precise differentiation and integration, vigilant tracking of signs and constants, and recognizing when the technique leads back to the original integral.