Integration by Parts from the Product Rule

Overview

Central question: How does reversing the product rule trade one integral for another? You will derive the rule, make a defensible choice of $u$ and $dv$, and verify that the trade actually simplified the problem. Integration by parts is not a command to split every product. It is useful only when differentiating one factor and integrating the other produces a more manageable remaining integral.

Start with a product you can differentiate easily: $x e^x$. The product rule gives

$$\frac{d}{dx}\left(xe^x\right)=xe^x+e^x.$$

Rearranging and integrating both sides yields

$$\int xe^x\,dx=xe^x-\int e^x\,dx.$$

This is the idea before it becomes a formula: one product integral is exchanged for another. The exchange helps here because $\int e^x\,dx$ is simpler than $\int xe^x\,dx$.

Derive the rule before using it

Let $u$ and $v$ be differentiable functions. The product rule is

$$d(uv)=u\,dv+v\,du.$$

Move the second term to the other side and integrate:

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

The symbols have jobs. Choose $u$ to be differentiated, so $du$ records its change. Choose $dv$ to be integrated, so $v$ is an antiderivative of $dv$. Do not treat $dv$ as an ordinary factor detached from its differential; the pair tells us which expression is being integrated.

Choose parts by measuring complexity

For $\int xe^x\,dx$, choose $u=x$ and $dv=e^x\,dx$. Then $du=dx$ and $v=e^x$, so

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

The choice works because differentiating $x$ removes its algebraic complexity while integrating $e^x$ leaves it unchanged. A common heuristic puts logarithms and inverse trigonometric functions ahead of algebraic factors, then trigonometric or exponential factors, but a heuristic is not a theorem. The proof of a good choice is the resulting integral: it must be simpler, solvable, or part of a useful cycle.

Contrast a poor split

Reverse the choices: $u=e^x$ and $dv=x\,dx$. Then $du=e^x\,dx$ and $v=\frac{x^2}{2}$, producing

$$\int xe^x\,dx =\frac{x^2e^x}{2}-\frac12\int x^2e^x\,dx.$$

The remaining integral is worse because the polynomial degree increased. This is not algebraically false; it is strategically poor. If an application of the rule makes the unknown integral more complicated, stop and choose differently.

Definite integrals use the same trade

Integration by parts also applies to a definite integral, provided the product term is evaluated at the endpoints. For example,

$$\int_0^1 xe^x\,dx =\left[xe^x\right]_0^1-\int_0^1e^x\,dx.$$

The first term is an endpoint contribution. At $x=1$ it is $e$; at $x=0$ it is zero. The remaining definite integral is $e-1$, so the complete value is $e-(e-1)=1$. Writing the brackets explicitly prevents an easy mistake: the product term is not an indefinite expression with a new constant; it is a definite difference of values.

This example also supplies a reasonableness check. On $0\le x\le1$, the integrand $xe^x$ is nonnegative, so the integral must be nonnegative. Its graph lies below the rectangle of height $e$ and width $1$, so the answer should be less than $e$. The value $1$ passes both checks. A graph or a rough numerical estimate cannot replace the calculation, but it can reveal a missing endpoint sign or an algebraic error.

A transfer example with a logarithm

The integral $\int\arctan(x)\,dx$ has no obvious product until it is written as $\int\arctan(x)\cdot1\,dx$. Choose $u=\arctan(x)$ and $dv=dx$. Then $du=\frac{1}{1+x^2}\,dx$ and $v=x$, producing

$$\int\arctan(x)\,dx =x\arctan(x)-\int\frac{x}{1+x^2}\,dx.$$

The remaining integral is now a substitution problem, not another parts problem. With $w=1+x^2$, $dw=2x\,dx$, so the answer is

$$x\arctan(x)-\frac12\ln\left(1+x^2\right)+C.$$

This example shows how techniques can cooperate. The first method is chosen from product structure; the second is chosen from a matching inside derivative. The handoff is justified by the new expression, not by a preselected sequence of formula names.

Stop and check

A plausible incorrect claim is: “LIATE says $u=x$, so the choice is proven.” The misconception is confusing a useful pattern with a mathematical justification. Targeted feedback: use the heuristic to propose a choice, then inspect the new integral. If it does not simplify, the heuristic has not licensed the calculation.

Retrieval check

Find an antiderivative of $\int \ln(x)\,dx$ for $x>0$. Choose $u=\ln(x)$ and $dv=dx$, giving $du=\frac1x\,dx$ and $v=x$. Therefore

$$\int\ln(x)\,dx=x\ln(x)-\int1\,dx=x\ln(x)-x+C.$$

Differentiate your answer. The Chain Rule reduces the derivative of $x\ln(x)$ to $\ln(x)+1$, and the derivative of $-x$ removes the extra $1$. The next lesson develops two special outcomes of the same trade: repeated reduction and a cycle that brings the original integral back.