In General Maths you integrated powers of x x x c c c calculus↺ ). Further Maths expressions need a little preparation before they can be integrated. This lesson collects the four ways to prepare them.
∫ a xⁿ dx = a xn + 1 (n + 1) + c
add 1 to the power , then divide by the new power ; never forget + c (n ≠ −1)
Integrating a power Add 1 to the power, divide by the new power, add c The curve from its gradient
Integrating a gradient gives the curve, up to the constant c c c c c c
Worked example · WAEC 2011
WAEC 2011 · Paper 2 · Q10 (a)
The gradient of a curve is given by 2 x − 3 x 2 2x - 3x^2 2 x − 3 x 2 ( 1 , 2 ) (1, 2) ( 1 , 2 )
Integrate the gradient
y = ∫ ( 2 x − 3 x 2 ) d x = x 2 − x 3 + c {y = \int (2x - 3x^2)\,dx = x^2 - x^3 + c} y = ∫ ( 2 x − 3 x 2 ) d x = x 2 − x 3 + c Think first. Integrate 2x − 3x².
Use the point
2 = 1 − 1 + c {2 = 1 - 1 + c} 2 = 1 − 1 + c c = 2 {c = 2} c = 2 The curve is y = x 2 − x 3 + 2 {y = x^2 - x^3 + 2} y = x 2 − x 3 + 2
Think first. Put x = 1 and y = 2.
← Back Next step → Show all steps Start again
Check yourself
The gradient of a curve is 6 x 2 − 4 x + 1 6x^2 - 4x + 1 6 x 2 − 4 x + 1 ( 1 , 5 ) (1, 5) ( 1 , 5 ) y = … y = \ldots y = …
More: the curve from its gradient
NECO 2023 · Paper 1 · Q17 A curve passes through ( − 2 , 3 ) (-2, 3) 3 x 2 + 8 x − 5 3x^2 + 8x - 5 WAEC 2014 · Paper 2 · Q9 Resolve x 2 + 1 ( x + 2 ) 3 \dfrac{x^2 + 1}{(x + 2)^3} WAEC 2017 · Paper 2 · Q11 If f ( x ) = ∫ ( 4 x − x 2 ) d x f(x) = \displaystyle\int(4x - x^2)\,dx f ( 3 ) = 21 f(3) = 21 f ( x ) f(x) WAEC 2008 · Paper 2 · Q9 A curve has gradient 3 x 2 − 2 x + 1 3x^2 - 2x + 1 ( x , y ) (x, y) ( 1 , 3 ) (1, 3) WAEC 2010 · Paper 2 · Q10 If x 2 + y 2 = p y ( 1 + x 2 ) x^2 + y^2 = py(1 + x^2) p p d y d x \dfrac{dy}{dx} WAEC 2022 · Paper 1 · Q34 The gradient of a function at any point ( x , y ) (x, y) 2 x − 6 2x - 6 ( 1 , 2 ) (1, 2) Simplify first
There is no product or quotient rule for integration. Before integrating, multiply out brackets and divide each term by a single power of x x x x x x
For example, x 3 − x 2 + 2 x 2 = x − 1 + 2 x − 2 {\frac{x^3 - x^2 + 2}{x^2} = x - 1 + 2x^{-2}} x 2 x 3 − x 2 + 2 = x − 1 + 2 x − 2
Then ∫ ( x − 1 + 2 x − 2 ) d x = x 2 2 − x − 2 x + c {\int (x - 1 + 2x^{-2})\,dx = \frac{x^2}{2} - x - \frac2x + c} ∫ ( x − 1 + 2 x − 2 ) d x = 2 x 2 − x − x 2 + c
More: simplify, then integrate
Substitution
When a bracket is raised to a power and multiplied by the derivative of what is inside, let u u u
∫ 2x (x² + 1 )³ dx: let u = x² + 1 , so du = 2x dx
= ∫ u³ du = u⁴ ⁄ 4 + c = (x² + 1)⁴ ⁄ 4 + c
let u be the inside; its derivative should appear outside; change the limits too
Substitution u = the inside; du = u′ dx; swap everything into u For a definite integral, change the limits to values of u u u
Integration by substitution Step through
∫ 3x²(x³ + 2)⁴ dx ∫ 3x²(x³ + 2)⁴ dx step 0 of 4
∫ 3x²(x³ + 2)⁴ dx ∫ x ÷ (x² + 5) dx ∫₀¹ 4x(2x² + 1)³ dx ← Back Next step →
A power of a bracket, multiplied by something. Look inside the bracket. Worked example · WAEC 2018
Choose u
Let u = 1 − x {u = 1 - x} u = 1 − x x = 1 − u {x = 1 - u} x = 1 − u d x = − d u {dx = -du} d x = − d u
Think first. What is inside the root? Write x and dx in terms of u.
Swap everything into u
∫ x 1 − x d x = ∫ ( 1 − u ) u 1 2 ( − d u ) {\int x\sqrt{1 - x}\,dx = \int (1 - u)u^{\frac12}(-du)} ∫ x 1 − x d x = ∫ ( 1 − u ) u 2 1 ( − d u ) = − ∫ ( u 1 2 − u 3 2 ) d u {= -\int \left(u^{\frac12} - u^{\frac32}\right)du} = − ∫ ( u 2 1 − u 2 3 ) d u Integrate
− ( 2 3 u 3 2 − 2 5 u 5 2 ) + c {-\left(\frac23u^{\frac32} - \frac25u^{\frac52}\right) + c} − ( 3 2 u 2 3 − 5 2 u 2 5 ) + c = 2 5 u 5 2 − 2 3 u 3 2 + c {= \frac25u^{\frac52} - \frac23u^{\frac32} + c} = 5 2 u 2 5 − 3 2 u 2 3 + c Think first. Add 1 to each power and divide by the new power.
Put x back
2 5 ( 1 − x ) 5 2 − 2 3 ( 1 − x ) 3 2 + c {\frac25(1 - x)^{\frac52} - \frac23(1 - x)^{\frac32} + c} 5 2 ( 1 − x ) 2 5 − 3 2 ( 1 − x ) 2 3 + c ← Back Next step → Show all steps Start again
Logarithms
When the top is the derivative of the bottom, the integral is a logarithm:
∫ 1 ⁄ (ax + b) dx = (1 ⁄ a) ln|ax + b| + c
∫ f′(x) ⁄ f(x) dx = ln|f(x) | + c
when the top is the derivative of the bottom, the answer is a log
Integrals that give logs ∫ f′(x) ÷ f(x) dx = ln|f(x)| + c Worked example · WAEC 2020
WAEC 2020 · Paper 2 · Q11 (b)
Evaluate ∫ − π 2 π 2 4 cos θ 3 + 2 sin θ d θ \displaystyle\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{4\cos\theta}{3 + 2\sin\theta}\,d\theta ∫ − 2 π 2 π 3 + 2 sin θ 4 cos θ d θ
Choose u
Let u = 3 + 2 sin θ {u = 3 + 2\sin\theta} u = 3 + 2 sin θ d u = 2 cos θ d θ {du = 2\cos\theta\,d\theta} d u = 2 cos θ d θ
The top, 4 cos θ d θ {4\cos\theta\,d\theta} 4 cos θ d θ 2 d u {2\,du} 2 d u
Think first. The bottom is 3 + 2 sin θ. What is its derivative?
Change the limits
θ = − π 2 {\theta = -\frac\pi2} θ = − 2 π u = 3 − 2 = 1 {u = 3 - 2 = 1} u = 3 − 2 = 1 θ = π 2 {\theta = \frac\pi2} θ = 2 π u = 3 + 2 = 5 {u = 3 + 2 = 5} u = 3 + 2 = 5 Think first. Work out u at θ = −π/2 and θ = π/2.
Integrate
∫ 1 5 2 u d u = [ 2 ln u ] 1 5 {\int_1^5 \frac2u\,du = [2\ln u]_1^5} ∫ 1 5 u 2 d u = [ 2 ln u ] 1 5 = 2 ln 5 − 2 ln 1 = 2 ln 5 ≈ 3.2189 {= 2\ln5 - 2\ln1 = 2\ln5 \approx 3.2189} = 2 ln 5 − 2 ln 1 = 2 ln 5 ≈ 3.2189 ← Back Next step → Show all steps Start again
Partial fractions first
A fraction with a factorised bottom can’t be integrated as it stands. Split it into partial fractions (see partial fractions↺ ), then integrate each piece: A x + a \frac{A}{x + a} x + a A B ( x + a ) 2 \frac{B}{(x + a)^2} ( x + a ) 2 B
Worked example · WAEC 2017
WAEC 2017 · Paper 2 · Q2
Evaluate ∫ 1 3 x − 1 ( x + 1 ) 2 d x \displaystyle\int_1^3 \frac{x - 1}{(x + 1)^2}\,dx ∫ 1 3 ( x + 1 ) 2 x − 1 d x
Partial fractions
x − 1 ( x + 1 ) 2 = A x + 1 + B ( x + 1 ) 2 {\frac{x - 1}{(x + 1)^2} = \frac{A}{x + 1} + \frac{B}{(x + 1)^2}} ( x + 1 ) 2 x − 1 = x + 1 A + ( x + 1 ) 2 B x − 1 = A ( x + 1 ) + B {x - 1 = A(x + 1) + B} x − 1 = A ( x + 1 ) + B Put x = − 1 x = -1 x = − 1 B = − 2 {B = -2} B = − 2 x x x A = 1 {A = 1} A = 1
Think first. Write x − 1 = A(x + 1) + B.
Integrate each piece
∫ 1 x + 1 d x = ln ( x + 1 ) {\int \frac{1}{x + 1}\,dx = \ln(x + 1)} ∫ x + 1 1 d x = ln ( x + 1 ) ∫ − 2 ( x + 1 ) − 2 d x = 2 x + 1 {\int -2(x + 1)^{-2}\,dx = \frac{2}{x + 1}} ∫ − 2 ( x + 1 ) − 2 d x = x + 1 2 So the integral is [ ln ( x + 1 ) + 2 x + 1 ] 1 3 {\left[\ln(x + 1) + \frac{2}{x + 1}\right]_1^3} [ ln ( x + 1 ) + x + 1 2 ] 1 3
The limits
At x = 3 x = 3 x = 3 ln 4 + 1 2 {\ln4 + \frac12} ln 4 + 2 1 x = 1 x = 1 x = 1 ln 2 + 1 {\ln2 + 1} ln 2 + 1
Subtract: ln 4 − ln 2 − 1 2 = ln 2 − 1 2 {\ln4 - \ln2 - \frac12 = \ln2 - \frac12} ln 4 − ln 2 − 2 1 = ln 2 − 2 1
≈ 0.6931 − 0.5 = 0.193 {\approx 0.6931 - 0.5 = 0.193} ≈ 0.6931 − 0.5 = 0.193 ← Back Next step → Show all steps Start again
Common mistake
Two slips to watch for:
Integrating a fraction by integrating the top and bottom separately. Simplify, substitute or split it first.
Forgetting + c +\,c + c
Your turn
(a) Express x + 6 ( x + 1 ) 3 \dfrac{x + 6}{(x + 1)^3} ( x + 1 ) 3 x + 6
(b) Use the answer in (a) to evaluate ∫ 1 2 x + 6 ( x + 1 ) 3 d x \displaystyle\int_1^2 \frac{x + 6}{(x + 1)^3}\,dx ∫ 1 2 ( x + 1 ) 3 x + 6 d x
Worked solution (try it first) (a) Write
A x + 1 + B ( x + 1 ) 2 + C ( x + 1 ) 3 \dfrac{A}{x + 1} + \dfrac{B}{(x + 1)^2} + \dfrac{C}{(x + 1)^3} x + 1 A + ( x + 1 ) 2 B + ( x + 1 ) 3 C and multiply through:
x + 6 = A ( x + 1 ) 2 + B ( x + 1 ) + C x + 6 = A(x + 1)^2 + B(x + 1) + C x + 6 = A ( x + 1 ) 2 + B ( x + 1 ) + C .
Put
x = − 1 x = -1 x = − 1 :
5 = C 5 = C 5 = C .
The
x 2 x^2 x 2 terms:
0 = A 0 = A 0 = A .
The
x x x terms:
1 = 2 A + B 1 = 2A + B 1 = 2 A + B , so
B = 1 B = 1 B = 1 .
So the answer is
1 ( x + 1 ) 2 + 5 ( x + 1 ) 3 \dfrac{1}{(x + 1)^2} + \dfrac{5}{(x + 1)^3} ( x + 1 ) 2 1 + ( x + 1 ) 3 5 .
(b) Integrate each piece:
∫ ( x + 1 ) − 2 d x = − 1 x + 1 \displaystyle\int (x + 1)^{-2}\,dx = -\frac{1}{x + 1} ∫ ( x + 1 ) − 2 d x = − x + 1 1 and
∫ 5 ( x + 1 ) − 3 d x = − 5 2 ( x + 1 ) 2 \displaystyle\int 5(x + 1)^{-3}\,dx = -\frac{5}{2(x + 1)^2} ∫ 5 ( x + 1 ) − 3 d x = − 2 ( x + 1 ) 2 5 .
At
x = 2 x = 2 x = 2 :
− 1 3 − 5 18 = − 11 18 -\frac13 - \frac{5}{18} = -\frac{11}{18} − 3 1 − 18 5 = − 18 11 .
At
x = 1 x = 1 x = 1 :
− 1 2 − 5 8 = − 9 8 -\frac12 - \frac58 = -\frac98 − 2 1 − 8 5 = − 8 9 .
Subtract:
− 11 18 + 9 8 = − 44 + 81 72 -\frac{11}{18} + \frac98 = \frac{-44 + 81}{72} − 18 11 + 8 9 = 72 − 44 + 81 = 37 72 = \frac{37}{72} = 72 37 ≈ 0.514 \approx 0.514 ≈ 0.514 .
Watch out
In (a), A = 0 A = 0 A = 0 x + 1 x + 1 x + 1 In (b), ∫ ( x + 1 ) − 3 d x = ( x + 1 ) − 2 − 2 \int (x + 1)^{-3}\,dx = \dfrac{(x + 1)^{-2}}{-2} ∫ ( x + 1 ) − 3 d x = − 2 ( x + 1 ) − 2 Report a problem with this question