Exam answer Show all explanations
Let I = ∫ ( x 3 + x + 1 ) ( x 2 − 1 ) d x I = \int\frac{(x^3 + x + 1)}{(x^2 - 1)}\,dx I = ∫ ( x 2 − 1 ) ( x 3 + x + 1 ) d x
Dividing, x 3 + x + 1 = x ( x 2 − 1 ) + ( 2 x + 1 ) x^3 + x + 1 = x(x^2 - 1) + (2x + 1) x 3 + x + 1 = x ( x 2 − 1 ) + ( 2 x + 1 )
Why?
The degree of the numerator (3) is not less than the degree of the denominator (2), so the fraction is improper. Divide first, then use partial fractions on the remainder.
x 3 ÷ x 2 = x x^3 \div x^2 = x x 3 ÷ x 2 = x x ( x 2 − 1 ) = x 3 − x x(x^2 - 1) = x^3 - x x ( x 2 − 1 ) = x 3 − x
( x 3 + x + 1 ) − ( x 3 − x ) = x 3 + x + 1 − x 3 + x = ( x 3 − x 3 ) + ( x + x ) + 1 = 2 x + 1 \begin{aligned} &(x^3 + x + 1) - (x^3 - x) \\ &= x^3 + x + 1 - x^3 + x \\ &= (x^3 - x^3) + (x + x) + 1 \\ &= 2x + 1 \end{aligned} ( x 3 + x + 1 ) − ( x 3 − x ) = x 3 + x + 1 − x 3 + x = ( x 3 − x 3 ) + ( x + x ) + 1 = 2 x + 1 Also x 2 − 1 = ( x − 1 ) ( x + 1 ) x^2 - 1 = (x - 1)(x + 1) x 2 − 1 = ( x − 1 ) ( x + 1 )
I = ∫ ( x + 2 x + 1 ( x − 1 ) ( x + 1 ) ) d x I = \int\left(x + \frac{2x + 1}{(x - 1)(x + 1)}\right)dx I = ∫ ( x + ( x − 1 ) ( x + 1 ) 2 x + 1 ) d x Now resolving into partial fractions, let
2 x + 1 ( x − 1 ) ( x + 1 ) = A x − 1 + B x + 1 \frac{2x + 1}{(x - 1)(x + 1)} = \frac{A}{x - 1} + \frac{B}{x + 1} ( x − 1 ) ( x + 1 ) 2 x + 1 = x − 1 A + x + 1 B Why?
The denominator has two different linear factors, so each one gets a constant over it.
⇒ 2 x + 1 = A ( x + 1 ) + B ( x − 1 ) \Rightarrow 2x + 1 = A(x + 1) + B(x - 1) ⇒ 2 x + 1 = A ( x + 1 ) + B ( x − 1 ) Why?
Multiply both sides by ( x − 1 ) ( x + 1 ) (x - 1)(x + 1) ( x − 1 ) ( x + 1 ) x x x x = 1 x = 1 x = 1 B B B x = − 1 x = -1 x = − 1 A A A
Putting x = 1 x = 1 x = 1 3 = 2 A ⇒ A = 3 2 3 = 2A \Rightarrow A = \frac32 3 = 2 A ⇒ A = 2 3
Putting x = − 1 x = -1 x = − 1 − 1 = − 2 B ⇒ B = 1 2 -1 = -2B \Rightarrow B = \frac12 − 1 = − 2 B ⇒ B = 2 1
Why?
Put each value into 2 x + 1 = A ( x + 1 ) + B ( x − 1 ) 2x + 1 = A(x + 1) + B(x - 1) 2 x + 1 = A ( x + 1 ) + B ( x − 1 )
x = 1 : 2 ( 1 ) + 1 = A ( 1 + 1 ) + B ( 1 − 1 ) 3 = 2 A + 0 A = 3 2 x = − 1 : 2 ( − 1 ) + 1 = A ( − 1 + 1 ) + B ( − 1 − 1 ) − 1 = 0 − 2 B B = − 1 − 2 = 1 2 \begin{aligned} x = 1&: 2(1) + 1 = A(1 + 1) + B(1 - 1) \\ &\ 3 = 2A + 0 \\ &\ A = \frac32 \\[4pt] x = -1&: 2(-1) + 1 \\ &\ = A(-1 + 1) + B(-1 - 1) \\ &\ -1 = 0 - 2B \\ &\ B = \frac{-1}{-2} = \frac12 \end{aligned} x = 1 x = − 1 : 2 ( 1 ) + 1 = A ( 1 + 1 ) + B ( 1 − 1 ) 3 = 2 A + 0 A = 2 3 : 2 ( − 1 ) + 1 = A ( − 1 + 1 ) + B ( − 1 − 1 ) − 1 = 0 − 2 B B = − 2 − 1 = 2 1 Then 3 / 2 x − 1 = 3 2 ( x − 1 ) \frac{3/2}{x - 1} = \frac{3}{2(x - 1)} x − 1 3/2 = 2 ( x − 1 ) 3 1 / 2 x + 1 = 1 2 ( x + 1 ) \frac{1/2}{x + 1} = \frac{1}{2(x + 1)} x + 1 1/2 = 2 ( x + 1 ) 1
∴ 2 x + 1 ( x − 1 ) ( x + 1 ) = 3 2 ( x − 1 ) + 1 2 ( x + 1 ) \begin{aligned} \therefore\ &\frac{2x + 1}{(x - 1)(x + 1)} \\ &= \frac{3}{2(x - 1)} + \frac{1}{2(x + 1)} \end{aligned} ∴ ( x − 1 ) ( x + 1 ) 2 x + 1 = 2 ( x − 1 ) 3 + 2 ( x + 1 ) 1 I = ∫ ( x + 3 2 ( x − 1 ) + 1 2 ( x + 1 ) ) d x = x 2 2 + 3 2 log ∣ x − 1 ∣ + 1 2 log ∣ x + 1 ∣ + C \begin{aligned} I &= \int\bigg(x + \frac{3}{2(x - 1)} \\ &\qquad + \frac{1}{2(x + 1)}\bigg)dx \\ &= \frac{x^2}{2} + \frac32\log|x - 1| \\ &\quad + \frac12\log|x + 1| + C \end{aligned} I = ∫ ( x + 2 ( x − 1 ) 3 + 2 ( x + 1 ) 1 ) d x = 2 x 2 + 2 3 log ∣ x − 1∣ + 2 1 log ∣ x + 1∣ + C Why?
Integrate term by term with the standard results
∫ x d x = x 2 2 ∫ d x x − a = log ∣ x − a ∣ \begin{aligned} \int x\,dx &= \frac{x^2}{2} \\ \int\frac{dx}{x - a} &= \log|x - a| \end{aligned} ∫ x d x ∫ x − a d x = 2 x 2 = log ∣ x − a ∣ Watch out: keep the modulus signs inside the logs.
To write the two logs as one, use k log p = log p k k\log p = \log p^k k log p = log p k log p + log q = log p q \log p + \log q = \log pq log p + log q = log pq
3 2 log ∣ x − 1 ∣ + 1 2 log ∣ x + 1 ∣ = 1 2 ( log ∣ x − 1 ∣ 3 + log ∣ x + 1 ∣ ) \begin{aligned} &\frac32\log|x - 1| + \frac12\log|x + 1| \\ &= \frac12\left(\log|x - 1|^3 + \log|x + 1|\right) \end{aligned} 2 3 log ∣ x − 1∣ + 2 1 log ∣ x + 1∣ = 2 1 ( log ∣ x − 1 ∣ 3 + log ∣ x + 1∣ ) In full:
3 2 log ∣ x − 1 ∣ = 1 2 ⋅ 3 log ∣ x − 1 ∣ = 1 2 log ∣ x − 1 ∣ 3 1 2 log ∣ x − 1 ∣ 3 + 1 2 log ∣ x + 1 ∣ = 1 2 ( log ∣ x − 1 ∣ 3 + log ∣ x + 1 ∣ ) = 1 2 log ( ∣ x − 1 ∣ 3 ∣ x + 1 ∣ ) = 1 2 log ∣ ( x − 1 ) 3 ( x + 1 ) ∣ \begin{aligned} &\frac32\log|x - 1| = \frac12 \cdot 3\log|x - 1| \\ &= \frac12\log|x - 1|^3 \\[4pt] &\frac12\log|x - 1|^3 + \frac12\log|x + 1| \\ &= \frac12\left(\log|x - 1|^3 + \log|x + 1|\right) \\ &= \frac12\log\left(|x - 1|^3\,|x + 1|\right) \\ &= \frac12\log\left|(x - 1)^3(x + 1)\right| \end{aligned} 2 3 log ∣ x − 1∣ = 2 1 ⋅ 3 log ∣ x − 1∣ = 2 1 log ∣ x − 1 ∣ 3 2 1 log ∣ x − 1 ∣ 3 + 2 1 log ∣ x + 1∣ = 2 1 ( log ∣ x − 1 ∣ 3 + log ∣ x + 1∣ ) = 2 1 log ( ∣ x − 1 ∣ 3 ∣ x + 1∣ ) = 2 1 log ( x − 1 ) 3 ( x + 1 ) ∴ \therefore ∴ I = x 2 2 + 1 2 log ∣ ( x − 1 ) 3 ( x + 1 ) ∣ + C I = \dfrac{x^2}{2} + \dfrac12\log\left|(x - 1)^3(x + 1)\right| + C I = 2 x 2 + 2 1 log ( x − 1 ) 3 ( x + 1 ) + C
Where this comes from Chapter 7, Integrals : Partial fractions after long division.
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 top x 3 + x + 1 x^3 + x + 1 x 3 + x + 1 x 2 − 1 x^2 - 1 x 2 − 1 x x x 2 x + 1 ( x − 1 ) ( x + 1 ) \frac{2x + 1}{(x - 1)(x + 1)} ( x − 1 ) ( x + 1 ) 2 x + 1 3 2 ( x − 1 ) + 1 2 ( x + 1 ) \frac{3}{2(x - 1)} + \frac{1}{2(x + 1)} 2 ( x − 1 ) 3 + 2 ( x + 1 ) 1
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