Topic 6.11 Integrating Using Integration by Parts: integrate products of functions by reversing the product rule, choosing the parts and applying the formula, including repeated use (BC).

What this topic is asking

The College Board (Topic 6.11, BC only) introduces integration by parts, the integration technique that reverses the product rule. When an integrand is a product of two functions of different kinds (for example a polynomial times an exponential, or a variable times a trigonometric function), substitution usually fails, and integration by parts is the tool that breaks the product apart.

The formula and where it comes from

Choosing the parts: LIATE

The whole skill is choosing $u$ so that the new integral $\int v\,du$ is easier than the one you started with. The mnemonic LIATE orders the function types by how good a choice they are for $u$: Logarithmic, Inverse trig, Algebraic (polynomial), Trigonometric, Exponential. Pick $u$ to be whichever factor comes first in that list. The logic is that logarithms and inverse trig functions are hard to integrate but easy to differentiate, so they make good $u$; exponentials and trig functions are easy to integrate, so they make good $dv$. Polynomials are good $u$ because each differentiation lowers the degree, eventually reaching a constant.

A worked single application

When the natural choice of dv is just dx

Some integrands look like a single function with nothing obvious to integrate, but parts still works by taking $dv = dx$. The standard case is $\int \ln x\,dx$: choose $u = \ln x$ and $dv = dx$, so $du = \frac{1}{x}\,dx$ and $v = x$. Then

The same trick handles $\int \arctan x\,dx$ and $\int \arcsin x\,dx$, where the inverse function is $u$ and $dv = dx$. Recognizing that $dv = dx$ is allowed is what makes these single-function integrals tractable.

Repeated integration by parts

When $u$ is a higher-degree polynomial, one application leaves an integral that still contains a product, so you apply parts again. For $\int x^2 e^x\,dx$, the first pass with $u = x^2$ leaves $\int 2x e^x\,dx$, which is the worked example above scaled by 2. Each pass lowers the polynomial degree by one until it disappears. A second family needs parts twice for a different reason: $\int e^x\sin x\,dx$ returns to a multiple of the original integral after two applications, and you solve for it algebraically. Writing $I = \int e^x\sin x\,dx$, two applications give $I = e^x\sin x - e^x\cos x - I$, so $2I = e^x\sin x - e^x\cos x$ and $I = \frac{1}{2}e^x(\sin x - \cos x) + C$.

A worked repeated application