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We have 2 x 2 + 3 x 2 ( x 2 + 9 ) \dfrac{2x^2 + 3}{x^2(x^2 + 9)} x 2 ( x 2 + 9 ) 2 x 2 + 3 x 2 = t x^2 = t x 2 = t
So, 2 t + 3 t ( t + 9 ) = A t + B t + 9 \text{So, }\frac{2t + 3}{t(t + 9)} = \frac{A}{t} + \frac{B}{t + 9} So, t ( t + 9 ) 2 t + 3 = t A + t + 9 B Why?
x x x x 2 x^2 x 2 t t t x 2 x^2 x 2 t t t t + 9 t + 9 t + 9
The letter t t t d x dx d x
⇒ 2 t + 3 = A ( t + 9 ) + B t \Rightarrow 2t + 3 = A(t + 9) + Bt ⇒ 2 t + 3 = A ( t + 9 ) + Bt Why?
Multiply both sides by t ( t + 9 ) t(t + 9) t ( t + 9 ) A t \frac{A}{t} t A A ( t + 9 ) A(t + 9) A ( t + 9 ) B t + 9 \frac{B}{t + 9} t + 9 B B t Bt Bt
Putting t = 0 t = 0 t = 0 3 = 9 A ⇒ A = 1 3 3 = 9A \Rightarrow A = \frac13 3 = 9 A ⇒ A = 3 1
Putting t = − 9 t = -9 t = − 9 − 15 = − 9 B ⇒ B = 5 3 -15 = -9B \Rightarrow B = \frac53 − 15 = − 9 B ⇒ B = 3 5
Why?
Choose values of t t t t = 0 t = 0 t = 0 B t Bt Bt t = − 9 t = -9 t = − 9 A ( t + 9 ) A(t + 9) A ( t + 9 )
At t = − 9 t = -9 t = − 9 2 ( − 9 ) + 3 = − 15 2(-9) + 3 = -15 2 ( − 9 ) + 3 = − 15
At t = 0 : 2 ( 0 ) + 3 = A ( 0 + 9 ) + B ( 0 ) 3 = 9 A A = 3 9 = 3 ÷ 3 9 ÷ 3 = 1 3 At t = − 9 : 2 ( − 9 ) + 3 = A ( − 9 + 9 ) + B ( − 9 ) − 18 + 3 = A ( 0 ) − 9 B − 15 = − 9 B B = − 15 − 9 = 15 9 = 15 ÷ 3 9 ÷ 3 = 5 3 \begin{aligned} &\text{At } t = 0: \\ 2(0) + 3 &= A(0 + 9) + B(0) \\ 3 &= 9A \\ A &= \frac39 = \frac{3 \div 3}{9 \div 3} = \frac13 \\[4pt] &\text{At } t = -9: \\ 2(-9) + 3 &= A(-9 + 9) + B(-9) \\ -18 + 3 &= A(0) - 9B \\ -15 &= -9B \\ B &= \frac{-15}{-9} = \frac{15}{9} \\ &= \frac{15 \div 3}{9 \div 3} = \frac53 \end{aligned} 2 ( 0 ) + 3 3 A 2 ( − 9 ) + 3 − 18 + 3 − 15 B At t = 0 : = A ( 0 + 9 ) + B ( 0 ) = 9 A = 9 3 = 9 ÷ 3 3 ÷ 3 = 3 1 At t = − 9 : = A ( − 9 + 9 ) + B ( − 9 ) = A ( 0 ) − 9 B = − 9 B = − 9 − 15 = 9 15 = 9 ÷ 3 15 ÷ 3 = 3 5 ∴ 2 x 2 + 3 x 2 ( x 2 + 9 ) = 1 3 x 2 + 5 3 ( x 2 + 9 ) \therefore\ \frac{2x^2 + 3}{x^2(x^2 + 9)} = \frac{1}{3x^2} + \frac{5}{3(x^2 + 9)} ∴ x 2 ( x 2 + 9 ) 2 x 2 + 3 = 3 x 2 1 + 3 ( x 2 + 9 ) 5 Why?
Put A = 1 3 A = \frac13 A = 3 1 B = 5 3 B = \frac53 B = 3 5 t = x 2 t = x^2 t = x 2
A t = 1 3 x 2 = 1 3 x 2 B t + 9 = 5 3 x 2 + 9 = 5 3 ( x 2 + 9 ) \begin{aligned} \frac{A}{t} &= \frac{\frac13}{x^2} = \frac{1}{3x^2} \\ \frac{B}{t + 9} &= \frac{\frac53}{x^2 + 9} = \frac{5}{3(x^2 + 9)} \end{aligned} t A t + 9 B = x 2 3 1 = 3 x 2 1 = x 2 + 9 3 5 = 3 ( x 2 + 9 ) 5 Using ∫ d x x 2 + a 2 = 1 a tan − 1 x a + c \int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + c ∫ x 2 + a 2 d x = a 1 tan − 1 a x + c
Why?
Put t = x 2 t = x^2 t = x 2 1 x 2 + 9 \frac{1}{x^2 + 9} x 2 + 9 1 a 2 = 9 a^2 = 9 a 2 = 9 a = 3 a = 3 a = 3
∫ 2 x 2 + 3 x 2 ( x 2 + 9 ) d x = 1 3 ∫ d x x 2 + 5 3 ∫ d x x 2 + 9 = − 1 3 x + 5 3 ⋅ 1 3 tan − 1 x 3 + c \begin{aligned} &\int\frac{2x^2 + 3}{x^2(x^2 + 9)}\,dx \\ &= \frac13\int\frac{dx}{x^2} + \frac53\int\frac{dx}{x^2 + 9} \\ &= -\frac{1}{3x} + \frac53\cdot\frac13\tan^{-1}\frac{x}{3} + c \end{aligned} ∫ x 2 ( x 2 + 9 ) 2 x 2 + 3 d x = 3 1 ∫ x 2 d x + 3 5 ∫ x 2 + 9 d x = − 3 x 1 + 3 5 ⋅ 3 1 tan − 1 3 x + c Why?
The first piece uses the power rule with n = − 2 n = -2 n = − 2
∫ x − 2 d x = x − 1 − 1 = − 1 x \int x^{-2}\,dx = \frac{x^{-1}}{-1} = -\frac1x ∫ x − 2 d x = − 1 x − 1 = − x 1 So 1 3 ∫ d x x 2 = − 1 3 x \frac13\int\frac{dx}{x^2} = -\frac{1}{3x} 3 1 ∫ x 2 d x = − 3 x 1 5 3 ⋅ 1 3 = 5 9 \frac53\cdot\frac13 = \frac59 3 5 ⋅ 3 1 = 9 5
1 3 ( − 1 x ) = − 1 3 x ∫ d x x 2 + 3 2 = 1 3 tan − 1 x 3 5 3 ⋅ 1 3 = 5 × 1 3 × 3 = 5 9 \begin{aligned} \frac13\left(-\frac1x\right) &= -\frac{1}{3x} \\[4pt] \int\frac{dx}{x^2 + 3^2} &= \frac13\tan^{-1}\frac{x}{3} \\ \frac53\cdot\frac13 &= \frac{5 \times 1}{3 \times 3} = \frac59 \end{aligned} 3 1 ( − x 1 ) ∫ x 2 + 3 2 d x 3 5 ⋅ 3 1 = − 3 x 1 = 3 1 tan − 1 3 x = 3 × 3 5 × 1 = 9 5 Watch out: ∫ d x x 2 \int\frac{dx}{x^2} ∫ x 2 d x log ∣ x 2 ∣ \log|x^2| log ∣ x 2 ∣ ∫ d x x \int\frac{dx}{x} ∫ x d x
= − 1 3 x + 5 9 tan − 1 ( x 3 ) + c = -\dfrac{1}{3x} + \dfrac59\tan^{-1}\left(\dfrac{x}{3}\right) + c = − 3 x 1 + 9 5 tan − 1 ( 3 x ) + c c c c
Where this comes from Chapter 7, Integrals : Integration by partial fractions (put x² = t).
Question type: Partial fractions
How to spot it: click to see Type 7. Partial fractions: break a fraction into simple pieces
Spot it: a fraction of two polynomials whose bottom splits into factors.
Here: The bottom x 2 ( x 2 + 9 ) x^2(x^2 + 9) x 2 ( x 2 + 9 ) x x x x 2 x^2 x 2 t t t x 2 x^2 x 2 2 t + 3 t ( t + 9 ) \frac{2t + 3}{t(t + 9)} t ( t + 9 ) 2 t + 3 1 3 t + 5 3 ( t + 9 ) \frac{1}{3t} + \frac{5}{3(t + 9)} 3 t 1 + 3 ( t + 9 ) 5 − 1 3 x -\frac{1}{3x} − 3 x 1 tan − 1 \tan^{-1} tan − 1
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