What is wrong LIATE choice?

Differentiating an exponential and integrating a logarithm leads nowhere; pick u by the LIATE order.

Use u = sin x to find sin^2 xcos x\,dx. [3 marks]

SQA AH style: by parts-style practice: Find x\,e^2x\,dx.

Integration by parts with u = x, dvdx = e^2x, so dudx = 1 and v = 12e^2x (1 mark for the choice). x e^2x\,dx = 12xe^2x - int 12e^2x\,dx (1 mark). = 12xe^2x - 14e^2x + C (2 marks). Markers reward the LIATE choice, the parts formula, and the second integration with + C.

SQA AH style: substitution-style practice: Use the substitution u = 1 + x^2 to find x(1 + x^2)^3\,dx.

With u = 1 + x^2, dudx = 2x, so x\,dx = 12\,du (1 mark). The integral becomes 1u^3·12\,du = 12int u^-3\,du (1 mark). = 12·u^-2-2 + C = -14u^2 + C (1 mark). Back-substitute: -14(1 + x^2)^2 + C (1 mark). Markers reward the substitution and du, the transformed integral, the power-rule result, and the back-substitution.

ScotlandMathsSyllabus dot point

Which technique do you reach for when an integral is not a standard result?

Integrate using standard results, integration by substitution, integration by parts and integration using partial fractions, and apply integration to find areas and volumes of revolution.

A focused answer to the SQA Advanced Higher Mathematics integration techniques content, covering standard results, integration by substitution, integration by parts, integration using partial fractions, and applications to areas and volumes of revolution.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-14

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

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What this dot point is asking
Standard results
Integration by substitution
Integration by parts
Integration using partial fractions
Areas and volumes of revolution
Try this

What this dot point is asking

The SQA wants you to integrate functions that are not standard results, by choosing the right technique: substitution, integration by parts, or a partial-fraction split. You also need to apply integration to find areas under curves and the volume of a solid of revolution.

Standard results

Know these by sight; they are the targets every technique aims to reach.

Integration by substitution

Substitution reverses the chain rule. Look for an inner function whose derivative also appears (up to a constant) in the integrand; call it $u$ , then rewrite everything in terms of $u$ and $du$ . For a definite integral, change the limits to $u$ -values so you never need to back-substitute.

Find

\displaystyle\int 2x\sqrt{x^2 + 1}\,dx

Substitution works when the integrand contains an inner function whose derivative also appears (up to a constant). Here, $x^2 + 1$ is inside the square root and its derivative $2x$ sits right next to it, making $u = x^2 + 1$ the natural choice.

Step 1: Choose the substitution and find $du$

Set $u = x^2 + 1$ . Differentiate to find $\dfrac{du}{dx} = 2x$ , which gives $du = 2x\,dx$ . The factor $2x\,dx$ in the integrand will be replaced by just $du$ .

Step 2: Rewrite the integral entirely in terms of $u$

Replace $x^2 + 1$ with $u$ and $2x\,dx$ with $du$ . The integral now involves only $u$ and is a standard power.

\int 2x\sqrt{x^2 + 1}\,dx = \int \sqrt{u}\,du = \int u^{1/2}\,du

Step 3: Integrate and back-substitute

Apply the power rule, then replace $u$ with $x^2 + 1$ to return to the original variable.

= \dfrac{u^{3/2}}{3/2} + C = \dfrac{2}{3}u^{3/2} + C

Final answer: $\dfrac{2}{3}(x^2 + 1)^{3/2} + C$ .

Integration by parts

Integration by parts reverses the product rule. The art is choosing which factor is $u$ (to be differentiated) and which is $dv$ (to be integrated). LIATE gives a reliable priority order for $u$ : logarithms, then inverse trig, then algebra, then trig, then exponentials.

Find

\displaystyle\int \ln x\,dx

At first glance this looks like a single function, not a product, so parts seems unnecessary. The trick is to write $\ln x = \ln x \cdot 1$ and set $dv = dx$ . By LIATE, the logarithm is the highest priority, so it becomes $u$ and gets differentiated, while the $1$ is integrated to give $v = x$ .

Step 1: Choose $u$ and $dv$ , then find $du$ and $v$

By LIATE, choose $u = \ln x$ (a logarithm) and $dv = dx$ (an algebraic constant, which integrates easily).

du = \dfrac{1}{x}\,dx, \qquad v = x

Step 2: Apply the parts formula $\int u\,dv = uv - \int v\,du$

Substitute the four pieces into the formula. The resulting integral is a simple one.

\int \ln x\,dx = x\ln x - \int x\cdot\dfrac{1}{x}\,dx = x\ln x - \int 1\,dx

Step 3: Complete the integration

The remaining integral is just $\int 1\,dx = x$ , so the result follows immediately.

= x\ln x - x + C

Final answer: $\displaystyle\int \ln x\,dx = x\ln x - x + C$ .

Integration using partial fractions

When the integrand is a rational function you cannot integrate directly, split it into partial fractions first; each piece is then a standard logarithm or power.

Find

\displaystyle\int \dfrac{1}{x^2 - 1}\,dx

The denominator factorises, so we split the integrand into partial fractions first. Each piece then becomes a standard logarithm integral that is straightforward to evaluate.

Step 1: Factorise and split into partial fractions

Write $x^2 - 1 = (x-1)(x+1)$ , then decompose using the cover-up rule: set $x = 1$ and $x = -1$ in turn to find $A$ and $B$ .

\dfrac{1}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}

At $x = 1$ : $A = \dfrac{1}{2}$ . At $x = -1$ : $B = -\dfrac{1}{2}$ .

Step 2: Integrate each term separately

Each partial fraction is of the form $\dfrac{k}{x - a}$ , which integrates to $k\ln|x - a|$ .

\int \dfrac{1}{x^2 - 1}\,dx = \int \dfrac{1/2}{x - 1}\,dx - \int \dfrac{1/2}{x + 1}\,dx = \dfrac{1}{2}\ln|x - 1| - \dfrac{1}{2}\ln|x + 1| + C

Step 3: Combine the logarithms using log laws

Use $\ln a - \ln b = \ln\dfrac{a}{b}$ to write the answer as a single logarithm.

= \dfrac{1}{2}\ln\left|\dfrac{x - 1}{x + 1}\right| + C

Final answer: $\displaystyle\int \dfrac{1}{x^2-1}\,dx = \dfrac{1}{2}\ln\left|\dfrac{x-1}{x+1}\right| + C$ .

Areas and volumes of revolution

The definite integral $\displaystyle\int_a^b y\,dx$ gives the signed area between a curve and the $x$ -axis. Rotating that region about the $x$ -axis produces a solid whose volume is found by summing thin discs of radius $y$ .

Try this

Q1. Find $\displaystyle\int xe^{x}\,dx$ . [3 marks]

Cue. Parts with $u = x$ : $xe^x - \displaystyle\int e^x\,dx = xe^x - e^x + C$ .

Q2. Use $u = \sin x$ to find $\displaystyle\int \sin^2 x\cos x\,dx$ . [3 marks]

Cue. $du = \cos x\,dx$ , so $\displaystyle\int u^2\,du = \dfrac{\sin^3 x}{3} + C$ .

Exam-style practice questions

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

AH style: by parts4 marksFind

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

Show worked answer →

Integration by parts with $u = x$ , $\dfrac{dv}{dx} = e^{2x}$ , so $\dfrac{du}{dx} = 1$ and $v = \dfrac{1}{2}e^{2x}$ (1 mark for the choice).

$\displaystyle\int x e^{2x}\,dx = \dfrac{1}{2}xe^{2x} - \int \dfrac{1}{2}e^{2x}\,dx$ (1 mark).

$= \dfrac{1}{2}xe^{2x} - \dfrac{1}{4}e^{2x} + C$ (2 marks). Markers reward the LIATE choice, the parts formula, and the second integration with $+ C$ .

AH style: substitution4 marksUse the substitution

u = 1 + x^2

to find

\displaystyle\int \dfrac{x}{(1 + x^2)^3}\,dx

Show worked answer →

With $u = 1 + x^2$ , $\dfrac{du}{dx} = 2x$ , so $x\,dx = \dfrac{1}{2}\,du$ (1 mark).

The integral becomes $\displaystyle\int \dfrac{1}{u^3}\cdot\dfrac{1}{2}\,du = \dfrac{1}{2}\int u^{-3}\,du$ (1 mark).

$= \dfrac{1}{2}\cdot\dfrac{u^{-2}}{-2} + C = -\dfrac{1}{4u^2} + C$ (1 mark).

Back-substitute: $-\dfrac{1}{4(1 + x^2)^2} + C$ (1 mark). Markers reward the substitution and $du$ , the transformed integral, the power-rule result, and the back-substitution.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)