Integrating Using Integration by Parts

students, imagine trying to find the total amount of water that flows into a tank when the flow rate is changing in a complicated way 📈. Sometimes the rate is not just a simple polynomial or exponential function. In AP Calculus BC, one powerful tool for finding accumulated change is integration by parts. It comes from the Product Rule for derivatives, and it helps us integrate products of functions that are hard to handle directly.

What integration by parts is and why it matters

Integration by parts is based on the derivative rule for products:

$$\frac{d}{dx}\big(u(x)v(x)\big)=u(x)v'(x)+v(x)u'(x).$$

If we integrate both sides, we get the main formula:

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

This formula lets us rewrite one difficult integral in terms of another, often simpler, integral. The idea is to choose part of the integrand as $u$ and the rest as $dv$. Then we compute $du$ and $v$, and substitute into the formula.

This technique is especially useful when you see a product like $x e^x$, $x\sin x$, $x\ln x$, or $\ln x$ by itself. These are common AP Calculus BC examples because they connect algebra, differentiation, and accumulation of change.

A helpful way to remember the formula is:

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

Think of it as “move one factor into the derivative, and the other into the antiderivative” ✍️.

Choosing $u$ and $dv$ wisely

The hardest part of integration by parts is often deciding what to let be $u$ and what to let be $dv$. A good choice usually makes the new integral easier than the original one.

A common guideline is to choose $u$ from a function that becomes simpler when differentiated. For example:

$u=\ln x$ becomes $\frac{1}{x}$, which is simpler.
$u=x^2$ becomes $2x$, which is simpler.
$u=\arctan x$ becomes $\frac{1}{1+x^2}$, which is simpler.

Then choose $dv$ as the remaining factor that is easy to integrate.

For example, in $\int x e^x\,dx$, let:

$$u=x, \quad dv=e^x\,dx.$$

Then:

$$du=dx, \quad v=e^x.$$

Now apply the formula:

$$\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C.$$

You can check this by differentiating the result. The derivative of $x e^x - e^x$ is $x e^x$, so the antiderivative is correct.

A classic example: integrating $x\ln x$

One of the most common AP Calculus BC examples is

$$\int x\ln x\,dx.$$

This is a product of algebra and logarithm. The logarithm becomes simpler when differentiated, so let:

$$u=\ln x, \quad dv=x\,dx.$$

Then:

$$du=\frac{1}{x}\,dx, \quad v=\frac{x^2}{2}.$$

Now use integration by parts:

$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \int \frac{x^2}{2}\cdot \frac{1}{x}\,dx.$$

Simplify the remaining integral:

$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{1}{2}\int x\,dx.$$

So:

$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C.$$

This example shows the big idea: integration by parts transforms a difficult product into a simpler polynomial integral.

Definite integrals and accumulated change

Integration by parts also works with definite integrals, which connect directly to accumulated change. If $f'(x)$ is a rate of change, then

$$\int_a^b f'(x)\,dx = f(b)-f(a).$$

When a product appears inside a definite integral, integration by parts can still rewrite the total accumulated change in a more useful way.

For example,

$$\int_0^1 x e^x\,dx.$$

Using the earlier setup with $u=x$ and $dv=e^x\,dx$, we get:

$$\int_0^1 x e^x\,dx = \Big[x e^x\Big]_0^1 - \int_0^1 e^x\,dx.$$

Now evaluate each part:

$$\Big[x e^x\Big]_0^1 = e,$$

and

$$\int_0^1 e^x\,dx = e-1.$$

So:

$$\int_0^1 x e^x\,dx = e-(e-1)=1.$$

This definite integral represents the total accumulation of the rate $x e^x$ from $x=0$ to $x=1$. In applied problems, that could model total distance, total work, or total amount accumulated over time depending on the context.

The connection to the Fundamental Theorem of Calculus

Integration by parts fits into AP Calculus BC because it works together with the Fundamental Theorem of Calculus. The theorem tells us how derivatives and definite integrals are related, while integration by parts helps us evaluate integrals that are not easy to do directly.

The formula

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

comes from combining the Product Rule with the idea of antiderivatives. In other words, it is another way to use derivative rules to solve accumulation problems.

This is important because many AP Calculus BC questions ask you to connect methods, not just calculate answers. For example, you might be asked to interpret what a definite integral means, choose an appropriate method, and then explain the result in context.

Repeated integration by parts and advanced cases

Sometimes one round of integration by parts is not enough. A famous example is

$$\int e^x\cos x\,dx.$$

A first application of integration by parts produces another integral of the same type. This can lead to an equation you solve for the original integral.

Let

$$u=\cos x, \quad dv=e^x\,dx.$$

Then

$$du=-\sin x\,dx, \quad v=e^x.$$

So:

$$\int e^x\cos x\,dx = e^x\cos x + \int e^x\sin x\,dx.$$

Now apply integration by parts again to $\int e^x\sin x\,dx$:

$$u=\sin x, \quad dv=e^x\,dx,$$

$$du=\cos x\,dx, \quad v=e^x.$$

Then:

$$\int e^x\sin x\,dx = e^x\sin x - \int e^x\cos x\,dx.$$

Substitute back:

$$\int e^x\cos x\,dx = e^x\cos x + e^x\sin x - \int e^x\cos x\,dx.$$

Now solve for the integral:

$$2\int e^x\cos x\,dx = e^x(\sin x+\cos x),$$

$$\int e^x\cos x\,dx = \frac{e^x(\sin x+\cos x)}{2}+C.$$

This kind of problem shows that integration by parts can be a strategic tool, not just a one-step formula.

When integration by parts is useful on the AP exam

students, integration by parts often appears when the integrand is a product of two functions and substitution is not the best first choice. It is especially useful when one factor becomes simpler after differentiation and the other is easy to integrate.

Typical patterns include:

$\int x e^x\,dx$
$\int x\sin x\,dx$
$\int x\cos x\,dx$
$\int \ln x\,dx$
$\int x^n\ln x\,dx$

You should also understand the meaning of the result in a real situation. If a rate function is integrated over time, the answer is the total accumulated change. If the integral is over an interval with units, the final units come from multiplying the units of the rate by the units of the variable of integration.

For example, if velocity is measured in feet per second and time is measured in seconds, then

$$\int_a^b v(t)\,dt$$

gives displacement in feet.

Conclusion

Integration by parts is a major AP Calculus BC technique for evaluating integrals of products. It comes directly from the Product Rule and fits naturally into the larger topic of accumulation of change. By choosing $u$ and $dv$ carefully, you can turn a difficult integral into a simpler one. This method works for indefinite and definite integrals, connects to the Fundamental Theorem of Calculus, and appears often in exam-style problems. Mastering it helps you solve problems that involve accumulated change in both pure math and real-world contexts 📚.

Study Notes

Integration by parts is based on the Product Rule:

$$\frac{d}{dx}(uv)=u\,v'+v\,u'$$

The main formula is:

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

Choose $u$ so that $du$ is simpler, and choose $dv$ so that $v$ is easy to find.
Common useful choices include $x$, $\ln x$, $\arctan x$, and inverse trigonometric functions as $u$.
A standard example is:

$$\int x e^x\,dx = x e^x - e^x + C$$

Another standard example is:

$$\int x\ln x\,dx = \frac{x^2}{2}\ln x - \frac{x^2}{4} + C$$

Integration by parts also works for definite integrals:

$$\int_a^b u\,dv = \Big[uv\Big]_a^b - \int_a^b v\,du$$

It is useful when one integration by parts step creates a simpler integral, or when repeating the process helps solve the original integral.
Integration by parts connects directly to accumulated change and the Fundamental Theorem of Calculus.
Always check your answer by differentiating when possible.