What is sign slip when v involves a trig function?

Integrating sin x gives -cos x; that minus then interacts with the formula's minus. Track signs carefully.

SACE Board SACE 2022-style practice: Calculator-free. Use integration by parts to find x e^2x\,dx.

Integration by parts: int u\,dv = uv - int v\,du. Choose u = x (so du = dx) and dv = e^2x\,dx (so v = 12e^2x), taking the algebraic factor as u. [1 mark] $int x e^2x\,dx = x·12e^2x - int 12e^2x\,dx = 12x e^2x - 12int e^2x\,dx. \;[1 mark]$ The remaining integral is 12e^2x, so = 12x e^2x - 14e^2x + c. [1 mark]

SACE Board SACE 2023-style practice: Calculator-free. Given x e^2x\,dx = 12x e^2x - 14e^2x + c, use integration by parts to show that x^2 e^2x\,dx = e^2x(12x^2 - 12x + 14) + k.

Choose u = x^2 (so du = 2x\,dx) and dv = e^2x\,dx (so v = 12e^2x). [1 mark] $int x^2 e^2x\,dx = 12x^2 e^2x - int 12e^2x(2x)\,dx = 12x^2 e^2x - int x e^2x\,dx. \;[1 mark]$ Substitute the given result: = 12x^2 e^2x - (12x e^2x - 14e^2x) + k = e^2x(12x^2 - 12x + 14) + k. [1 mark]

SACE Board SACE 2021-style practice: Calculator-free. Find ln x\,dx.

There is no obvious product, so take u = ln x and dv = dx, giving du = 1x\,dx and v = x: $int ln x\,dx = xln x - int x·1x\,dx = xln x - int 1\,dx = xln x - x + C.$ Marks: one for the choice u = ln x, dv = dx, one for the answer xln x - x + C.

§-Syllabus dot point

SASpecialist MathematicsSyllabus dot point

How does reversing the product rule let us integrate products such as x times e to the x?

Evaluate integrals using integration by parts, including choosing u and dv and applying the method more than once where needed.

Integration by parts as the reverse of the product rule, choosing u and dv with the LIATE guide, repeated application, and the standard trick for integrating the natural logarithm.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The formula
Choosing u and dv
A first example
Why the choice of u matters
Integrating the logarithm
Applying the method twice
The cyclic case

What this dot point is asking

You need to evaluate integrals using integration by parts, choosing $u$ and $dv$ sensibly and applying the method repeatedly when required.

The formula

Starting from the product rule $\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$ and integrating both sides gives

Choosing u and dv

A first example

Integration by parts once

Find $\displaystyle\int x e^{x}\,dx$

Step 1: Choose $u$ and $dv$ using LIATE

LIATE lists Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential, and the type that appears first should be chosen as $u$ . Here the algebraic term $x$ ranks higher than the exponential $e^x$ , so we set $u = x$ and $dv = e^{x}\,dx$ . This choice works because differentiating $x$ reduces it to a constant, making the remaining integral easier.

Step 2: Find $du$ and $v$

Differentiate $u$ and integrate $dv$ to get the pieces needed for the formula.

du = dx, \qquad v = \int e^{x}\,dx = e^{x}.

Step 3: Apply the formula and simplify

Substituting into $\int u\,dv = uv - \int v\,du$ and evaluating the remaining integral gives the answer.

\int x e^{x}\,dx = uv - \int v\,du = x e^{x} - \int e^{x}\,dx = x e^{x} - e^{x} + C = e^{x}(x - 1) + C.

Final answer: $e^{x}(x - 1) + C$

Why the choice of u matters

The formula trades the original integral $\int u\,dv$ for a new one $\int v\,du$ . The method only helps if that new integral is easier, which happens when $u$ becomes simpler on differentiation. A power of $x$ differentiates toward a lower power and eventually a constant, so it is usually a good $u$ when paired with an exponential or trig $dv$ . A logarithm or inverse trig function, by contrast, has no elementary antiderivative of its own but a tidy derivative, so it should almost always be chosen as $u$ . The LIATE order encodes exactly this reasoning, which is why following it so reliably produces a simpler second integral.

Integrating the logarithm

A classic case is $\int \ln x\,dx$ , where there is seemingly no product. Take $u = \ln x$ and $dv = dx$ :

\int \ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

Applying the method twice

When the algebraic factor is a higher power, one pass reduces it but leaves another integral by parts.

Applying integration by parts twice

Find $\displaystyle\int x^{2}\sin x\,dx$

Step 1: First application, choose $u = x^2$

By LIATE the algebraic $x^2$ ranks above the trigonometric $\sin x$ , so set $u = x^2$ and $dv = \sin x\,dx$ . Differentiating and integrating gives $du = 2x\,dx$ and $v = -\cos x$ .

Step 2: Apply the formula to get the remaining integral

Substituting into the integration by parts formula reduces the power of $x$ by one, but the result still contains a product that needs a second pass.

\int x^2\sin x\,dx = -x^2\cos x - \int (-\cos x)(2x)\,dx = -x^2\cos x + 2\int x\cos x\,dx.

Step 3: Second application, integrate $\int x\cos x\,dx$

Apply integration by parts again with $u = x$ and $dv = \cos x\,dx$ , giving $du = dx$ and $v = \sin x$ . Now the remaining integral is a single trig function, which is straightforward.

\int x\cos x\,dx = x\sin x - \int \sin x\,dx = x\sin x + \cos x.

Step 4: Substitute back and collect terms

Replace $\int x\cos x\,dx$ in the result from Step 2 with the expression found in Step 3, then simplify.

\int x^2\sin x\,dx = -x^2\cos x + 2\big(x\sin x + \cos x\big) + C = -x^2\cos x + 2x\sin x + 2\cos x + C.

The cyclic case

A special situation arises with integrals such as $\int e^x\cos x\,dx$ , where applying integration by parts twice returns a multiple of the original integral. Rather than continuing forever, you set the original integral equal to the expression you have derived, treat it as an unknown, and solve algebraically. For example, two passes give $I = e^x\sin x + e^x\cos x - I$ , so $2I = e^x(\sin x + \cos x)$ and $I = \tfrac{1}{2}e^x(\sin x + \cos x) + C$ . Recognising when the integral reappears, and solving for it instead of integrating again, is the trick this type of question rewards.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20223 marksCalculator-free. Use integration by parts to find

\displaystyle\int x e^{2x}\,dx

Show worked answer →

Integration by parts: $\int u\,dv = uv - \int v\,du$ .

Choose $u = x$ (so $du = dx$ ) and $dv = e^{2x}\,dx$ (so $v = \tfrac{1}{2}e^{2x}$ ), taking the algebraic factor as $u$ . [1 mark]

\int x e^{2x}\,dx = x\cdot\tfrac{1}{2}e^{2x} - \int \tfrac{1}{2}e^{2x}\,dx = \tfrac{1}{2}x e^{2x} - \tfrac{1}{2}\int e^{2x}\,dx. \;[\text{1 mark}]

The remaining integral is $\tfrac{1}{2}e^{2x}$ , so $= \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + c$ . [1 mark]

SACE 20233 marksCalculator-free. Given

\displaystyle\int x e^{2x}\,dx = \tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x} + c

, use integration by parts to show that

\displaystyle\int x^2 e^{2x}\,dx = e^{2x}\left(\tfrac{1}{2}x^2 - \tfrac{1}{2}x + \tfrac{1}{4}\right) + k

Show worked answer →

Choose $u = x^2$ (so $du = 2x\,dx$ ) and $dv = e^{2x}\,dx$ (so $v = \tfrac{1}{2}e^{2x}$ ). [1 mark]

\int x^2 e^{2x}\,dx = \tfrac{1}{2}x^2 e^{2x} - \int \tfrac{1}{2}e^{2x}(2x)\,dx = \tfrac{1}{2}x^2 e^{2x} - \int x e^{2x}\,dx. \;[\text{1 mark}]

Substitute the given result: $= \tfrac{1}{2}x^2 e^{2x} - \left(\tfrac{1}{2}x e^{2x} - \tfrac{1}{4}e^{2x}\right) + k = e^{2x}\left(\tfrac{1}{2}x^2 - \tfrac{1}{2}x + \tfrac{1}{4}\right) + k$ . [1 mark]

SACE 20212 marksCalculator-free. Find

\displaystyle\int \ln x\,dx

Show worked answer →

There is no obvious product, so take $u = \ln x$ and $dv = dx$ , giving $du = \tfrac{1}{x}\,dx$ and $v = x$ :

\int \ln x\,dx = x\ln x - \int x\cdot\tfrac{1}{x}\,dx = x\ln x - \int 1\,dx = x\ln x - x + C.

Marks: one for the choice $u = \ln x$ , $dv = dx$ , one for the answer $x\ln x - x + C$ .