6.11 Integrating Using Integration by Parts (BC only)

Some integrals become easy only after you reverse a familiar differentiation pattern: the product rule leads to integration by parts, while a complicated rational expression can be split into simpler fractions through linear partial fractions. Both techniques are designated BC only.

Integration by parts: reversing the product rule

The product rule says

$$ \frac{d}{dx}\bigl[u(x)v(x)\bigr]

u(x)\frac{dv}{dx} + v(x)\frac{du}{dx}. $$

Rearranging its differential form and integrating gives the integration-by-parts formula:

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

The method is useful when an integrand is a product whose factors behave differently under differentiation and integration. Usually, choose $u$ as the factor that becomes simpler when differentiated, and choose $dv$ as the factor that can be integrated readily.

A practical choice pattern is:

• Let $u$ be an algebraic factor such as $x$, $x^2$, or $\ln x$.
• Let $dv$ contain a factor with an immediate antiderivative, such as $e^x,dx$, $\sin x,dx$, or $\cos x,dx$.
• Compute $du$ by differentiating $u$ and $v$ by integrating $dv$.
• Substitute into $uv-\int v,du$.
• Differentiate the result to verify it.

Worked example: $\int x e^x,dx$

Choose

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

Then

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

Applying integration by parts,

$$ \int xe^x,dx

xe^x-\int e^x,dx

xe^x-e^x+C. $$

Therefore,

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

Verification uses differentiation:

$$ \frac{d}{dx}\left[e^x(x-1)\right]

e^x(x-1)+e^x

xe^x. $$

The derivative reproduces the original integrand, so the antiderivative is correct.

Misconception check — “Either factor works equally well.”
Choosing $u=e^x$ and $dv=x,dx$ is possible, but it produces an integral involving $x^2e^x$, which is more complicated rather than simpler. Integration by parts is not merely a formula-substitution exercise; the choice of $u$ and $dv$ determines whether the remaining integral improves.

The two recognition patterns

The methods can be distinguished by the structure of the integrand:

Integrand structure	Recognition cue	Main move
Product such as $x e^x$	Reverse product rule	$\int u,dv=uv-\int v,du$
Rational function with distinct linear factors	Split the denominator into linear pieces	Partial-fraction decomposition
Ratio with a linear denominator	Numerator is related to the denominator’s derivative	Reverse chain rule and logarithm

For FUN-6.E, integration by parts is used to determine indefinite or definite integrals when a product structure calls for it. For FUN-6.F, linear partial fractions are used to determine indefinite or definite integrals after decomposition. The associated suggested skill is 1.E — Apply appropriate mathematical rules or procedures, with and without technology, under Mathematical Practice 1: Implementing Mathematical Processes.

Retrieval check: For

$$ \int \frac{7}{5x+1},dx, $$

identify the substitution and the coefficient multiplying the logarithm. Then decide whether

$$ \int x\cos x,dx $$

is better suited to substitution or integration by parts. The answers are $u=5x+1$, coefficient $\frac{7}{5}$, and integration by parts.