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Techniques of integration

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By the end, you can

  • Recognise reverse chain-rule patterns
  • Use substitution and integration by parts
  • Check antiderivatives by differentiating

about 35 minutes for the lesson · practise at your own pace.

The central idea

The printed outline covers f′(x)[f(x)]nf^{\prime}(x)[f(x)]^n (including n=−1n=-1), f′(x)ef(x)f^{\prime}(x)e^{f(x)}, sin⁡2x\sin^2x, cos⁡2x\cos^2x, tan⁡2x\tan^2x, 1/(a2+x2)1/(a^2+x^2), 1/a2−x21/\sqrt{a^2-x^2}, 1/(a2−x2)1/(a^2-x^2) and 1/(x2−a2)1/(x^2-a^2), followed by a given substitution and integration by parts. The methods below connect these forms to the worked examples and practice routes.

Integration and differentiation undo each other

If F′(x)=f(x)F^{\prime}(x)=f(x), then ∫f(x) dx=F(x)+C\int f(x)\,dx=F(x)+C; ff is the derivative and F+CF+C is the indefinite integral. For example, ddxx3=3x2\frac{d}{dx}x^3=3x^2 is equivalent to ∫3x2 dx=x3+C\int3x^2\,dx=x^3+C.

For continuous f,gf,g and a constant kk, use these basic rules:

RulePrinted example
∫(f±g) dx=∫f dx±∫g dx\int(f\pm g)\,dx=\int f\,dx\pm\int g\,dx∫(x3+3x2) dx=∫x3 dx+∫3x2 dx\int(x^3+3x^2)\,dx=\int x^3\,dx+\int3x^2\,dx
∫kf dx=k∫f dx\int kf\,dx=k\int f\,dx∫3x3 dx=3∫x3 dx\int3x^3\,dx=3\int x^3\,dx
∫ddxf(x) dx=f(x)+C\int\frac{d}{dx}f(x)\,dx=f(x)+C∫ddx(3x2) dx=∫6x dx=3x2+C\int\frac{d}{dx}(3x^2)\,dx=\int6x\,dx=3x^2+C
ddx[∫f(x) dx]=f(x)\frac{d}{dx}[\int f(x)\,dx]=f(x)ddx[∫3x2 dx]=ddx(x3+C)=3x2\frac{d}{dx}[\int3x^2\,dx]=\frac{d}{dx}(x^3+C)=3x^2

Use one arbitrary constant for a resulting family on each connected interval. The Teacher highlights continuity and the two inverse-operation rules.

An indefinite integral describes antiderivatives, so include a constant. A substitution changes both the expression and its differential. Integration by parts, ∫u dv=uv−∫v du\int u\,dv=uv-\int v\,du, reverses the product rule. Trigonometric identities and partial fractions can simplify an integrand before either method.

Reverse the chain rule

Differentiating (5x2+2x+1)8(5x^2+2x+1)^8 gives 8(5x2+2x+1)7(10x+2)=16(5x+1)(5x2+2x+1)78(5x^2+2x+1)^7(10x+2)=16(5x+1)(5x^2+2x+1)^7. Hence

∫(5x+1)(5x2+2x+1)7 dx=116(5x2+2x+1)8+C.\int(5x+1)(5x^2+2x+1)^7\,dx=\frac1{16}(5x^2+2x+1)^8+C.

More generally, reverse the derivative of [f(x)]n+1[f(x)]^{n+1}:

∫f′(x)[f(x)]n dx=[f(x)]n+1n+1+C,n≠−1.\int f^{\prime}(x)[f(x)]^n\,dx=\frac{[f(x)]^{n+1}}{n+1}+C,\qquad n\ne-1.

Dividing an arbitrary constant by nonzero n+1 leaves an arbitrary constant, so rename it C. The handwritten strategy says to try standard forms first, then use additional techniques if they do not fit. It labels the basic power rule O-level, the composite rule H2-level and this general form not in MF27. Match the inner derivative before applying the rule.

Explore

Compare xex2xe^{x^2} with xe2xxe^{2x}. Before integrating, differentiate each exponent and decide whether its derivative accounts for the remaining factor. Try substitution u=x2u=x^2 in the first and parts with u=xu=x in the second. Explain what each choice simplifies, then differentiate both proposed answers. A familiar-looking product alone does not determine the method.

Related interactive example

Explore this related example, then return to the prediction task above.

Does (ln x)⁶/12 track the area?

Move the two bounds to select an interval. Compare its signed integral with the change in the antiderivative.

Integration boundsa = 1.00, x = 2.50

Move either handle to choose the interval. The graphs compare the signed integral with F(x) − F(a).

Try an example

How the shaded areas add up

The limits run left to right. Area above the axis adds; area below the axis subtracts.

Above-axis contribution
0.049
Below-axis contribution
0
Signed integral
0.049

All three bars share one scale. Positive contributions extend right of zero; negative contributions extend left.

Signed integral from 1.00 to x
0.049
F(x) − F(a)
0.049
agrees at displayed precision
Formula − integral
0
0.00% of the integral’s magnitude
Compare the running integral on a graph
  • Integrand
  • Numerical signed integral from the start to x
  • Part of the integrand above the axis
  • Part of the integrand below the axis
  • F(x) − F(start), from the antiderivative
Key idea

Differentiating (ln⁡x)612\frac{(\ln x)^6}{12} gives (ln⁡x)52x\frac{(\ln x)^5}{2x}: the dashed curve and the solid curve agree.

Ideas to try
  1. Move both bounds below 1: the integrand and signed integral are negative.
  2. Set the lower bound to 1 and the upper bound to ee: what is the integral?
Open the full interactive problem →

Worked example

∫2x(x2+1)3 dx\int2x(x^2+1)^3\,dx: let u=x2+1u=x^2+1, so du=2x dxdu=2x\,dx. The result is u4/4+C=(x2+1)4/4+Cu^4/4+C=(x^2+1)^4/4+C.

Substituting the expression but leaving the old differential mixes variables.

Check your understanding

Using the given substitution u=x2+1u=x^2+1, find ∫xln⁡(x2+1) dx\int x\ln(x^2+1)\,dx. Explain the next method needed after substitution, give the real domain and verify the final answer by differentiation.

Mastery learning objectives

At the end of this chapter, I should be able to:

Recognise general standard forms

Match f′(x)[f(x)]nf^{\prime}(x)[f(x)]^n, n≠−1n\ne-1.

Match f′(x)/f(x)f^{\prime}(x)/f(x) and its logarithmic antiderivative.

Match f′(x)ef(x)f^{\prime}(x)e^{f(x)} or f′(x)af(x)f^{\prime}(x)a^{f(x)}.

Recognise 1/(x2+a2)1/(x^2+a^2), 1/a2−x21/\sqrt{a^2-x^2}, 1/(x2−a2)1/(x^2-a^2) and 1/(a2−x2)1/(a^2-x^2).

Recognise trigonometric standard forms

Match f′(x)cos⁡[f(x)]f^{\prime}(x)\cos[f(x)].

Match f′(x)sin⁡[f(x)]f^{\prime}(x)\sin[f(x)].

Match f′(x)sec⁡2[f(x)]f^{\prime}(x)\sec^2[f(x)].

Match f′(x)csc⁡2[f(x)]f^{\prime}(x)\csc^2[f(x)].

Match f′(x)sec⁡[f(x)]tan⁡[f(x)]f^{\prime}(x)\sec[f(x)]\tan[f(x)].

Match f′(x)csc⁡[f(x)]cot⁡[f(x)]f^{\prime}(x)\csc[f(x)]\cot[f(x)].

Transform to a suitable form

Complete the square.

Use partial fractions.

Split the numerator.

Use trigonometric identities and formulas.

Use substitution.

Use integration by parts.

Choose a practice question

Common question typeLecture examplesDiscussionReview follow-up
Integration as anti-differentiation11, 2—
Composite power, n≠−1n\ne-123(a)–(d), 6(a)–(b)11(b)
Logarithmic derivative33(e)–(g), 6(c)11(a)
Composite exponential/base-a43(h)–(i)6(b), 10(i)
Inverse-trig and difference-of-squares forms5, 6, 7411(c)
Partial fractions / split numerator8, 9, 1053(a), 12, 13
Trigonometric standard forms116(c)11(e)
Trigonometric identities12, 13, 14, 156(d)–(e), 122(i)
Substitution167, 11(ii)2(ii), 4, 6(a), 8(a), 9, 10(ii), 15(ii), as printed
Parts17, 188, 9, 10, 11(i)1, 3(b), 5, 8(b), 11(d), 14, 15(i), as printed

The printed power-map condition n≠1 is corrected to n≠−1 here, consistent with the preceding derivation. The printed map refers to Review15(i)/(ii), although Review15has one ask; the two-method parts/substitution comparison is Review16(i)–(iii). These source references remain visible rather than being silently reassigned. The map is a selected practice route; omitted questions/subparts still retain their full canonical records.

Two printed classifications also need care: Discussion6(c) appears in both the logarithmic and trigonometric rows, but its working uses the cosecant-squared reverse chain rule; Review14 appears under parts, but its requested method is partial fractions. Follow the linked working for the actual method.

Practice questions

Practice section 1 · Questions 1–6

Reverse chain rules, powers, logarithms and exponentials

Practice section 2 · Questions 7–12

Exponential, inverse-trigonometric and logarithmic standard forms

Practice section 3 · Questions 13–18

Partial fractions, denominator derivatives and trigonometric composites

Practice section 4 · Questions 19–24

Trigonometric identities and substitution

Practice section 5 · Questions 25–30

Integration by parts, given derivatives and direct integrals

Practice section 6 · Questions 31–36

Mixed methods: standard forms, partial fractions, trigonometry, substitution and parts

Practice section 7 · Questions 37–42

Combining given derivatives, trigonometric identities and parts

Practice section 8 · Questions 43–48

Substitution, trigonometric identities and derivative-assisted parts

Practice section 9 · Questions 49–54

Choosing between standard forms, partial fractions and repeated parts

Practice section 10 · Question 55

Compare integration by parts with an arctangent substitution

More practice

Explain why your chosen method simplifies the integral, track the variable and domain, and differentiate the final expression before accepting it.

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For “Techniques of integration”, try a question without the working and explain your method. This records your confidence, saved in this browser only.

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