Integration by substitution, including substitutions that are given to you in harder cases

Changing the limits of a definite integral when you substitute

Integration by parts, including ∫ln x dx and products with ln x

Using integration by parts twice, keeping control of the signs

Integrating rational functions by first writing them in partial fractions, including repeated factors and improper fractions

Recognising which method an integral needs

Chaining two methods in one longer question

1. When no standard result fits, there are three tools: substitution rewrites the integral in a simpler variable, integration by parts trades a product for an easier integral, and partial fractions turn a rational function into terms that integrate to logarithms.

2. Substitution: choose u=g(x), replace dx using du/dx, and rewrite everything in u before integrating. Return to x, or, for a definite integral, change the limits to u-values and never go back.

3. By parts: ∫udv/dx dx=uv-∫vdu/dx dx. Let u be the factor that gets simpler when differentiated, but if there is a ln x, let u=ln x.

4. ln x on its own: write it as 1×ln x, so ∫ln x dx=xln x-x+c.

5. With x² times eˣ, sin x or cos x, the first use of parts leaves a power of x; use parts again.

6. Partial fractions: each linear term gives k/aln|ax+b|; a squared factor gives -k/(a(ax+b)). Divide first if the fraction is improper.

Substitution is the chain rule run backwards. If an integral contains a function g(x) and also (a multiple of) its derivative, writing u=g(x) collapses it to an integral in u that you already know. Even when the derivative is not obviously there, a well-chosen u can turn an awkward expression, such as a power of x-3 multiplied by x, into a sum of simple powers.

The method has three steps. (1) Choose u (or use the one given) and differentiate to link du and dx. (2) Replace every part of the integrand and the dx; if any x remains, the substitution is not finished.

(3) Integrate with respect to u, then substitute back to give the answer in x.

WORKED EXAMPLE · QUESTION

(a) Use the substitution u=x-3 to find ∫x(x-3)⁵ dx, giving your answer in the form k(x-3)⁶(px+q)+c. (b) Use the substitution u=eˣ+1 to find ∫e²ˣ/(eˣ+1) dx.

SOLUTION

(a) du/dx=1, so dx=du, and x=u+3. Then ∫(u+3)u⁵ du=∫(u⁶+3u⁵) du=u⁷/7+u⁶/2+c. Back to x and factorise out (1/14)(x-3)⁶: (1/14)(x-3)⁶[2(x-3)+7]= (1/14)(x-3)⁶(2x+1)+c.

(b) du/dx=eˣ, so eˣ dx=du. The numerator e²ˣ=eˣ×eˣ supplies du with one eˣ=u-1 left over. So the integral is ∫(u-1)/u du=∫(1-1/u)du=u-ln u+c. Back to x: eˣ+1-ln(eˣ+1)+c, which is eˣ-ln(eˣ+1)+c.

In (b) the "+1" has been absorbed into the arbitrary constant, and ln u needs no modulus because u=eˣ+1>0. Check by differentiating: eˣ-eˣ/(eˣ+1)=e²ˣ/(eˣ+1) ✓.

The limits of a definite integral are values of the variable you integrate with respect to. When the variable changes from x to u, the limits must change too: put each x-limit into u=g(x). Then evaluate in u; there is no need to go back to x. Geometrically, the substitution changes the shape of the region but not its area, as the two figures below show.

WORKED EXAMPLE · QUESTION

Use the substitution u=2x-1 to show that ∫₁⁵x/(√(2x-1)) dx=(16/3).

SOLUTION

du/dx=2, so dx=½ du, and x=(u+1)/2. New limits: x=1⇒u=1; x=5⇒u=9.

Rewrite entirely in u: ∫₁⁹((u+1)/2)/√u×½ du=¼∫₁⁹(u^(1/2)+u^(-1/2))du.

Integrate and use the u-limits: ¼[⅔u^(3/2)+2u^(1/2)]₁⁹=¼[(18+6)-(⅔+2)]=¼×(64/3)= (16/3) as required.

Integrating the product rule, d/dx(uv)=udv/dx+vdu/dx, and rearranging gives a formula for integrating a product.

Three rules cover almost every case. A power of x times eˣ, sin or cos: u= the power of x, because differentiating lowers the power while eˣ, sin and cos are easy to integrate. Anything times ln x: u=ln x, because you cannot easily integrate ln x but its derivative is 1/x. And ln x on its own: write it as 1×ln x, with u=ln x and dv/dx=1.

WORKED EXAMPLE · QUESTION

Find ∫xcos 2x dx. Hence find the area between y=xcos 2x and the x-axis from x=0 to x=π/4.

SOLUTION

Choose u=x (differentiates to 1) and dv/dx=cos 2x: du/dx=1, v=½sin 2x. Then ∫xcos 2x dx=½xsin 2x-∫½sin 2x dx= ½xsin 2x+¼cos 2x+c.

Area: [½xsin 2x+¼cos 2x]₀^(π/4)=(π/8+0)-(0+¼)= π/8-¼≈0.143.

Choosing u=cos 2x instead would make v=½x²: the new integral, of ½x²×(-2sin 2x), is harder than the original.

WORKED EXAMPLE · QUESTION

(a) Find ∫ln x dx. (b) Hence show that ∫₁ᵉln x dx=1. (c) Find the exact value of ∫₁ᵉ xln x dx.

SOLUTION

(a) Write ln x=1×ln x; u=ln x, dv/dx=1, so du/dx=1/x, v=x. Then ∫ln x dx=xln x-∫x×1/x dx= xln x-x+c.

(b) With ln e=1 and ln 1=0: [xln x-x]₁ᵉ=(e-e)-(0-1)=1 as required.

(c) u=ln x again, now with dv/dx=x, so v=½x²: ∫xln x dx=½x²ln x-∫½x²×1/x dx=½x²ln x-¼x²+c. Evaluate: (½e²-¼e²)-(0-¼)= (e²+1)/4.

Each use of parts with u=xⁿ lowers the power by one. With x² the first application leaves an integral containing x, so parts is needed a second time. Keep careful track of signs: the second application sits inside a minus sign from the first.

WORKED EXAMPLE · QUESTION

Find the exact value of ∫₀^(π/2)x²sin x dx.

SOLUTION

First application, u=x², dv/dx=sin x, v=-cos x: ∫x²sin x dx=-x²cos x+∫2xcos x dx.

Second application, u=2x, dv/dx=cos x, v=sin x: ∫2xcos x dx=2xsin x-∫2sin x dx=2xsin x+2cos x.

Combine: ∫x²sin x dx=-x²cos x+2xsin x+2cos x+c.

Limits, with cos π/2=0, sin π/2=1, cos 0=1: (0+π+0)-(0+0+2)= π-2.

A rational function such as (5x+4)/((x-1)(x+2)) has no standard integral, but its partial fractions do. Split it first (the algebra is on the P4 partial fractions page), then integrate each term.

With distinct linear factors each term gives a logarithm; combine them with the laws of logarithms when an exact answer "in the form ln k" is asked for. A repeated factor (x+b)² needs both A/(x+b) and B/((x+b)²), and the second integrates to a fraction, not a log. An improper fraction (degree of numerator ≥ degree of denominator) must be divided first: for example (x²+x+1)/(x(x+1))=1+1/x-1/(x+1), so its integral is x+ln|x|-ln|x+1|+c.

WORKED EXAMPLE · QUESTION

Show that ∫₂⁴(5x+4)/((x-1)(x+2)) dx=ln k, stating the exact value of k.

SOLUTION

Partial fractions: (5x+4)/((x-1)(x+2))≡A/(x-1)+B/(x+2), so 5x+4≡A(x+2)+B(x-1). x=1: 9=3A⇒A=3. x=-2: -6=-3B⇒B=2.

Integrate: ∫₂⁴(3/(x-1)+2/(x+2))dx=[3ln|x-1|+2ln|x+2|]₂⁴.

Evaluate: (3ln 3+2ln 6)-(3ln 1+2ln 4)=3ln 3+2ln(3/2). Combine: ln 27+ln(9/4)=ln(243/4), so k=(243/4).

WORKED EXAMPLE · QUESTION

Given that (x+5)/((x+1)²(x+3))≡A/(x+1)+B/((x+1)²)+C/(x+3), find A, B and C, and hence find the exact value of ∫₀¹(x+5)/((x+1)²(x+3)) dx.

SOLUTION

x+5≡A(x+1)(x+3)+B(x+3)+C(x+1)². x=-1: 4=2B⇒B=2. x=-3: 2=4C⇒C=½. Compare x²: 0=A+C⇒A=-½.

Integrate term by term; the middle term is a power, not a log: [-½ln(x+1)-2/(x+1)+½ln(x+3)]₀¹.

Evaluate: (-½ln 2-1+½ln 4)-(0-2+½ln 3)=1+½ln 2-½ln 3= 1+½ln⅔ (≈0.797).

½ln 4=ln 2, which is how -½ln 2+½ln 4 becomes +½ln 2. Equivalent exact forms, such as 1-½ln 1.5, are equally correct.

Before anything else, look for a standard integral or a "recognition" integral (f′/f, or a function times its inside's derivative), which can be written down in one line. Then ask what kind of integrand you have.

Longer questions often chain two techniques: a substitution turns the integrand into a rational function of u, which then needs partial fractions. Keep each stage on its own lines (limits, the u-integral, the partial fractions, the evaluation) so that each can be followed.

WORKED EXAMPLE · QUESTION

I=∫₀^(ln 2)6eˣ/((eˣ+1)(eˣ+4)) dx. (a) Use the substitution u=eˣ to show that I=∫₁²6/((u+1)(u+4)) du. (b) Express 6/((u+1)(u+4)) in partial fractions. (c) Hence show that I=ln(25/16).

SOLUTION

(a) du/dx=eˣ, so eˣ dx=du. Limits: x=0⇒u=1; x=ln 2⇒u=2. So I=∫₁²6/((u+1)(u+4)) du as required.

(b) 6≡A(u+4)+B(u+1). u=-1: 6=3A⇒A=2. u=-4: 6=-3B⇒B=-2. So 2/(u+1)-2/(u+4).

(c) [2ln(u+1)-2ln(u+4)]₁²=(2ln 3-2ln 6)-(2ln 2-2ln 5)=2ln(3×5)/(6×2)=2ln(5/4)= ln(25/16) (≈0.446).

P3.7 Integration of standard functions and P3.6 differentiation

Every technique here ends in a standard result from P3: eˣ, 1/x, sin, cos, f(ax+b) and f′/f. Substitution is the chain rule reversed; parts is the product rule reversed.

P4.1 Partial fractions, P4.6 Volumes of revolution, P4.7 Differential equations

The algebra of splitting fractions comes first; areas, volumes of revolution and separable differential equations then need exactly these integrals, often as the second half of a longer question. Partial fractions are also used to expand rational functions as binomial series.

Mechanics and physics

Work done by a variable force, and displacement from a variable acceleration, lead to integrals such as ∫te⁻ᵏᵗ dt, which need integration by parts; exponential decay and capacitor discharge give similar integrals.

1. Find ∫xe⁻ˣ dx.

2. Use the substitution u=1+x² to find ∫₀¹x/((1+x²)²) dx.

3. Which method would you use for ∫1/(x²-1) dx, and what is the result?

4. In ∫xln x dx, why is u=ln x the right choice rather than u=x?

5. Find the exact value of ∫₁²ln x dx.

6. Find ∫(x²+x+1)/(x(x+1)) dx.

1. Parts with u=x, v=-e⁻ˣ: -xe⁻ˣ+∫e⁻ˣ dx=-xe⁻ˣ-e⁻ˣ+c=-(x+1)e⁻ˣ+c.

2. du=2x dx; limits 1 and 2. ∫₁²1/2u² du=½[-1/u]₁²=½(-½+1)=¼.

3. Partial fractions: 1/(x²-1)=(1/2)/(x-1)-(1/2)/(x+1), so the integral is ½ln|x-1|-½ln|x+1|+c=½ln|(x-1)/(x+1)|+c.

4. With u=x you would need v=∫ln x dx, itself a parts integral. With u=ln x, du/dx=1/x cancels against v=½x², leaving the easy ∫½x dx.

5. [xln x-x]₁²=(2ln 2-2)-(0-1)=2ln 2-1.

6. The fraction is improper, so divide first: (x²+x+1)/(x(x+1))=1+1/(x(x+1))=1+1/x-1/(x+1). The integral is x+ln|x|-ln|x+1|+c.