Exam answer Show all explanations
Put x 2 = y x^2 = y x 2 = y
Why?
x x x x 2 x^2 x 2 x 2 x^2 x 2 y y y
Watch out: this is not a substitution in the integral. d x dx d x y y y
x 2 + 1 ( x 2 + 2 ) ( x 2 + 3 ) = y + 1 ( y + 2 ) ( y + 3 ) = A y + 2 + B y + 3 \begin{aligned} &\frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)} = \frac{y + 1}{(y + 2)(y + 3)} \\ &= \frac{A}{y + 2} + \frac{B}{y + 3} \end{aligned} ( x 2 + 2 ) ( x 2 + 3 ) x 2 + 1 = ( y + 2 ) ( y + 3 ) y + 1 = y + 2 A + y + 3 B ⇒ y + 1 = A ( y + 3 ) + B ( y + 2 ) … ( i ) \begin{aligned} \Rightarrow y + 1 &= A(y + 3) \\ &\quad + B(y + 2) \quad\ldots(\text{i}) \end{aligned} ⇒ y + 1 = A ( y + 3 ) + B ( y + 2 ) … ( i ) Why?
Multiply both sides by ( y + 2 ) ( y + 3 ) (y + 2)(y + 3) ( y + 2 ) ( y + 3 ) y y y
Comparing coefficients of y y y
A + B = 1 and 3 A + 2 B = 1 A + B = 1 \text{ and } 3A + 2B = 1 A + B = 1 and 3 A + 2 B = 1 Why?
The right side of (i) is ( A + B ) y + ( 3 A + 2 B ) (A + B)y + (3A + 2B) ( A + B ) y + ( 3 A + 2 B ) 1 ⋅ y + 1 1 \cdot y + 1 1 ⋅ y + 1
A ( y + 3 ) + B ( y + 2 ) = A y + 3 A + B y + 2 B = ( A + B ) y + ( 3 A + 2 B ) \begin{aligned} &A(y + 3) + B(y + 2) \\ &= Ay + 3A + By + 2B \\ &= (A + B)y + (3A + 2B) \end{aligned} A ( y + 3 ) + B ( y + 2 ) = A y + 3 A + B y + 2 B = ( A + B ) y + ( 3 A + 2 B ) A quicker way: put y = − 2 y = -2 y = − 2 − 1 = A -1 = A − 1 = A y = − 3 y = -3 y = − 3 − 2 = − B -2 = -B − 2 = − B B = 2 B = 2 B = 2
y = − 2 : − 2 + 1 = A ( 1 ) + B ( 0 ) − 1 = A y = − 3 : − 3 + 1 = A ( 0 ) + B ( − 1 ) − 2 = − B , B = 2 \begin{aligned} y = -2&: -2 + 1 = A(1) + B(0) \\ &\ \ -1 = A \\[4pt] y = -3&: -3 + 1 = A(0) + B(-1) \\ &\ \ -2 = -B,\ B = 2 \end{aligned} y = − 2 y = − 3 : − 2 + 1 = A ( 1 ) + B ( 0 ) − 1 = A : − 3 + 1 = A ( 0 ) + B ( − 1 ) − 2 = − B , B = 2 Putting B = 1 − A B = 1 - A B = 1 − A
3 A + 2 ( 1 − A ) = 1 ⇒ A + 2 = 1 3A + 2(1 - A) = 1 \Rightarrow A + 2 = 1 3 A + 2 ( 1 − A ) = 1 ⇒ A + 2 = 1 Solving, A = − 1 A = -1 A = − 1 B = 2 B = 2 B = 2
Why?
From A + B = 1 A + B = 1 A + B = 1 B = 1 − A B = 1 - A B = 1 − A 3 A + 2 B = 1 3A + 2B = 1 3 A + 2 B = 1
3 A + 2 ( 1 − A ) = 1 3 A + 2 − 2 A = 1 A + 2 = 1 A = 1 − 2 = − 1 B = 1 − A = 1 − ( − 1 ) = 1 + 1 = 2 \begin{aligned} 3A + 2(1 - A) &= 1 \\ 3A + 2 - 2A &= 1 \\ A + 2 &= 1 \\ A &= 1 - 2 = -1 \\[4pt] B &= 1 - A = 1 - (-1) \\ &= 1 + 1 = 2 \end{aligned} 3 A + 2 ( 1 − A ) 3 A + 2 − 2 A A + 2 A B = 1 = 1 = 1 = 1 − 2 = − 1 = 1 − A = 1 − ( − 1 ) = 1 + 1 = 2 ∫ x 2 + 1 ( x 2 + 2 ) ( x 2 + 3 ) d x = ∫ − 1 x 2 + 2 d x + 2 ∫ 1 x 2 + 3 d x \begin{aligned} &\int\frac{x^2 + 1}{(x^2 + 2)(x^2 + 3)}\,dx \\ &= \int\frac{-1}{x^2 + 2}\,dx + 2\int\frac{1}{x^2 + 3}\,dx \end{aligned} ∫ ( x 2 + 2 ) ( x 2 + 3 ) x 2 + 1 d x = ∫ x 2 + 2 − 1 d x + 2 ∫ x 2 + 3 1 d x Why?
Put A = − 1 A = -1 A = − 1 B = 2 B = 2 B = 2 x 2 x^2 x 2 y y y
Using ∫ d x x 2 + a 2 = 1 a tan − 1 x a + C \displaystyle\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
x 2 + 2 = x 2 + ( 2 ) 2 x 2 + 3 = x 2 + ( 3 ) 2 \begin{aligned} x^2 + 2 &= x^2 + (\sqrt2)^2 \\ x^2 + 3 &= x^2 + (\sqrt3)^2 \end{aligned} x 2 + 2 x 2 + 3 = x 2 + ( 2 ) 2 = x 2 + ( 3 ) 2 Why?
To use the standard integral, write each constant as a square: a = 2 a = \sqrt2 a = 2 a = 3 a = \sqrt3 a = 3
So the first gives − 1 2 tan − 1 x 2 -\frac{1}{\sqrt2}\tan^{-1}\frac{x}{\sqrt2} − 2 1 tan − 1 2 x 2 × 1 3 tan − 1 x 3 2 \times \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} 2 × 3 1 tan − 1 3 x C C C
− ∫ d x x 2 + ( 2 ) 2 = − 1 2 tan − 1 x 2 2 ∫ d x x 2 + ( 3 ) 2 = 2 ⋅ 1 3 tan − 1 x 3 = 2 3 tan − 1 x 3 \begin{aligned} -\int\frac{dx}{x^2 + (\sqrt2)^2} &= -\frac{1}{\sqrt2}\tan^{-1}\frac{x}{\sqrt2} \\[4pt] 2\int\frac{dx}{x^2 + (\sqrt3)^2} &= 2 \cdot \frac{1}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} \\ &= \frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3} \end{aligned} − ∫ x 2 + ( 2 ) 2 d x 2 ∫ x 2 + ( 3 ) 2 d x = − 2 1 tan − 1 2 x = 2 ⋅ 3 1 tan − 1 3 x = 3 2 tan − 1 3 x ∫ x 2 + 1 ( x 2 + 2 ) ( x 2 + 3 ) d x \int\dfrac{x^2 + 1}{(x^2 + 2)(x^2 + 3)}\,dx ∫ ( x 2 + 2 ) ( x 2 + 3 ) x 2 + 1 d x = − 1 2 tan − 1 ( x 2 ) = -\dfrac{1}{\sqrt2}\tan^{-1}\left(\dfrac{x}{\sqrt2}\right) = − 2 1 tan − 1 ( 2 x ) + 2 3 tan − 1 ( x 3 ) + C + \dfrac{2}{\sqrt3}\tan^{-1}\left(\dfrac{x}{\sqrt3}\right) + C + 3 2 tan − 1 ( 3 x ) + C
Where this comes from Chapter 7, Integrals : Partial fractions in x².
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 + 2 ) ( x 2 + 3 ) (x^2 + 2)(x^2 + 3) ( x 2 + 2 ) ( x 2 + 3 ) x x x x 2 x^2 x 2 y y y x 2 x^2 x 2 − 1 y + 2 + 2 y + 3 -\frac{1}{y + 2} + \frac{2}{y + 3} − y + 2 1 + y + 3 2 tan − 1 \tan^{-1} tan − 1
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