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Chapter 7 · Integrals

Exercise 7.5 · Q23 (MCQ): ∫dxx(x2+1)\int\frac{dx}{x(x^2 + 1)}

Exercise 7.5Integration by partial fractions · Section 7.5
The question

Choose the correct answer.

∫dxx(x2+1)  equals\int\frac{dx}{x(x^2 + 1)} \ \text{ equals}

(A) log⁡∣x∣−12log⁡(x2+1)+C\log|x| - \frac12\log(x^2 + 1) + C
(B) log⁡∣x∣+12log⁡(x2+1)+C\log|x| + \frac12\log(x^2 + 1) + C
(C) −log⁡∣x∣+12log⁡(x2+1)+C-\log|x| + \frac12\log(x^2 + 1) + C
(D) 12log⁡∣x∣+log⁡(x2+1)+C\frac12\log|x| + \log(x^2 + 1) + C

Exam answer

Let II be the given integral.

1x(x2+1)=(x2+1)−x2x(x2+1)=1x−xx2+1\begin{aligned} \frac{1}{x(x^2 + 1)} &= \frac{(x^2 + 1) - x^2}{x(x^2 + 1)} \\ &= \frac1x - \frac{x}{x^2 + 1} \end{aligned} ∴ I=∫dxx−∫x dxx2+1=log⁡∣x∣−12log⁡(x2+1)+C\begin{aligned} \therefore\ I &= \int\frac{dx}{x} - \int\frac{x\,dx}{x^2 + 1} \\ &= \log|x| - \frac12\log(x^2 + 1) + C \end{aligned}

Answer: (A) log⁡∣x∣−12log⁡(x2+1)+C\log|x| - \dfrac12\log(x^2 + 1) + C

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The denominator is a linear factor xx times the irreducible quadratic x2+1x^2 + 1⁠, so Table 7.2 (row 5) gives Ax+Bx+Cx2+1\frac{A}{x} + \frac{Bx + C}{x^2 + 1}⁠. The constants turn out to be A=1A = 1⁠, B=−1B = -1⁠, C=0C = 0⁠, so the integral is log⁡∣x∣−12log⁡(x2+1)+C\log|x| - \frac12\log(x^2 + 1) + C⁠: option (A).

Given, to find, and the plan
Given
The integrand 1x(x2+1)\frac{1}{x(x^2 + 1)}⁠, defined for x≠0x \ne 0⁠, and four options.
To find
Which option is ∫dxx(x2+1)\int\frac{dx}{x(x^2 + 1)}⁠.
Plan
(1) Split into partial fractions. (2) Integrate. (3) Differentiate back. (4) Test every option by differentiating it.

Basics you need

1Table 7.2, row 5

px2+qx+r(x−a)(x2+bx+c)=Ax−a+Bx+Cx2+bx+c\frac{px^2 + qx + r}{(x - a)(x^2 + bx + c)} = \frac{A}{x - a} + \frac{Bx + C}{x^2 + bx + c} when the quadratic cannot be factorised. Lesson: partial fractions

2∫x dxx2+1\int\frac{x\,dx}{x^2 + 1}

Put t=x2+1t = x^2 + 1⁠, dt=2x dxdt = 2x\,dx⁠: the integral is 12log⁡(x2+1)+C\frac12\log(x^2 + 1) + C⁠. Basics: substitution

3Testing an option

An option is correct exactly when its derivative equals the integrand. Basics: integration undoes differentiation

Solution

  1. Split into partial fractions

    The fraction is proper (degree 0 over degree 3), and x2+1x^2 + 1 has no real roots. So

    1x(x2+1)=Ax+Bx+Cx2+1,\frac{1}{x(x^2 + 1)} = \frac{A}{x} + \frac{Bx + C}{x^2 + 1}, 1=A(x2+1)+(Bx+C)x=(A+B)x2+Cx+A.\begin{aligned} 1 &= A(x^2 + 1) + (Bx + C)x \\ &= (A + B)x^2 + Cx + A. \end{aligned}

    Compare coefficients: constant A=1A = 1⁠; xx⁠: C=0C = 0⁠; x2x^2⁠: A+B=0A + B = 0⁠, so B=−1B = -1⁠. Hence

    1x(x2+1)=1x−xx2+1.\frac{1}{x(x^2 + 1)} = \frac1x - \frac{x}{x^2 + 1}.
    Row 5 of Table 7.2 with a=0a = 0⁠, b=0b = 0⁠, c=1c = 1⁠. The cleared equation is an identity, so the coefficients of each power of xx must match those of 1=0x2+0x+11 = 0x^2 + 0x + 1⁠.
    Table 7.2, row 5. Basics: splitting fractions

  2. Integrate

    ∫dxx=log⁡∣x∣+C1,∫x dxx2+1=12log⁡(x2+1)+C2.\begin{aligned} \int\frac{dx}{x} &= \log|x| + C_1, \\ \int\frac{x\,dx}{x^2 + 1} &= \frac12\log(x^2 + 1) + C_2. \end{aligned}

    So

    ∫dxx(x2+1)=log⁡∣x∣−12log⁡(x2+1)+C.\int\frac{dx}{x(x^2 + 1)} = \log|x| - \frac12\log(x^2 + 1) + C.
    The first is standard formula (xii). For the second, t=x2+1t = x^2 + 1 gives x dx=12dtx\,dx = \frac12 dt and 12∫dtt\frac12\int\frac{dt}{t}⁠; since x2+1>0x^2 + 1 > 0 no modulus is needed. This is option (A).
    Standard formula (xii), Section 7.2; substitution, Section 7.3.1. Basics: logarithms and 1/x

  3. Check: differentiate the answer

    ddx(log⁡∣x∣−12log⁡(x2+1))=1x−xx2+1=(x2+1)−x2x(x2+1)=1x(x2+1). ✓\begin{aligned} &\frac{d}{dx}\Big(\log|x| - \tfrac12\log(x^2 + 1)\Big) \\ &= \frac1x - \frac{x}{x^2 + 1} \\ &= \frac{(x^2 + 1) - x^2}{x(x^2 + 1)} = \frac{1}{x(x^2 + 1)}. \ \checkmark \end{aligned}
    Chain rule: ddx12log⁡(x2+1)=12⋅2xx2+1\frac{d}{dx}\frac12\log(x^2 + 1) = \frac12\cdot\frac{2x}{x^2 + 1}⁠. Over the common denominator the x2x^2 terms cancel and 1 remains.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation

  4. Examine every option

    At x=1x = 1 the integrand is 11⋅2=0.5\frac{1}{1\cdot 2} = 0.5⁠.

    • (A) Its derivative is the integrand (Step 3). Correct.
    • (B) Derivative 1x+xx2+1=2x2+1x(x2+1)\frac1x + \frac{x}{x^2 + 1} = \frac{2x^2 + 1}{x(x^2 + 1)}⁠. At x=1x = 1⁠: 1.51.5⁠. The sign of the second log is wrong. Wrong.
    • (C) Derivative −1x+xx2+1=−1x(x2+1)-\frac1x + \frac{x}{x^2 + 1} = \frac{-1}{x(x^2 + 1)}⁠, exactly the negative of the integrand. At x=1x = 1⁠: −0.5-0.5⁠. Wrong.
    • (D) Derivative 12x+2xx2+1=5x2+12x(x2+1)\frac{1}{2x} + \frac{2x}{x^2 + 1} = \frac{5x^2 + 1}{2x(x^2 + 1)}⁠. At x=1x = 1⁠: 1.51.5⁠. Wrong.
    An option is an antiderivative exactly when its derivative equals the integrand for all x≠0x \ne 0⁠. Written over the denominator x(x2+1)x(x^2 + 1)⁠, the derivatives of (B), (C), (D) have numerators 2x2+12x^2 + 1⁠, −1-1 and 5x2+12\frac{5x^2 + 1}{2}⁠, none identically equal to 1; one sample point (x=1x = 1⁠) where they differ is enough to reject each.
Result ∫dxx(x2+1)=log⁡∣x∣−12log⁡(x2+1)+C,\int\frac{dx}{x(x^2 + 1)} = \log|x| - \frac12\log(x^2 + 1) + C,

for x≠0x \ne 0⁠. The correct option is (A). (Textbook answer: A.)

Check it with numbers

Decomposition at x=2x = 2⁠: 12⋅5=0.1\frac{1}{2\cdot 5} = 0.1 and 12−25=0.1\frac12 - \frac25 = 0.1⁠. ✓

Definite integral from 1 to 2 (continuous there): F(2)−F(1)=log⁡2−12log⁡5+12log⁡2F(2) - F(1) = \log 2 - \frac12\log 5 + \frac12\log 2 =12log⁡85≈0.2350018= \frac12\log\frac85 \approx 0.2350018⁠. A numerical integration over [1,2][1, 2] gives 0.23500180.2350018 too. ✓

Common mistakes
  1. Choosing (C). Its derivative is −1x(x2+1)-\frac{1}{x(x^2 + 1)}⁠: every sign is reversed.
  2. Using Bx2+1\frac{B}{x^2 + 1} with a constant BB⁠. Then the x2x^2 and constant equations cannot both hold; the quadratic factor needs Bx+CBx + C⁠.
  3. ∫x dxx2+1=log⁡(x2+1)\int\frac{x\,dx}{x^2 + 1} = \log(x^2 + 1)⁠. The derivative of x2+1x^2 + 1 is 2x2x⁠, so a factor 12\frac12 is needed.
What to take away
  1. 1x(x2+1)=1x−xx2+1\frac{1}{x(x^2 + 1)} = \frac1x - \frac{x}{x^2 + 1} (check: (x2+1)−x2=1(x^2 + 1) - x^2 = 1⁠).
  2. This is Exercise 7.5 Q16 with n=2n = 2⁠: 12log⁡x2x2+1\frac12\log\frac{x^2}{x^2 + 1} is the same answer.
  3. Differentiating each option settles an integration MCQ.

Every fact used, and where it comes from

FactUsed inSource
Ax−a+Bx+Cx2+bx+c\frac{A}{x - a} + \frac{Bx + C}{x^2 + bx + c}Step 1Table 7.2, row 5, Section 7.5
Equal polynomials have equal coefficientsStep 1Algebra; Remark after Example 11
∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 2Standard formula (xii), Section 7.2
Substitution t=x2+1t = x^2 + 1Step 2Section 7.3.1 (Lesson: substitution)
F′=fF' = f tests an antiderivativeSteps 3, 4Property (I), Section 7.2.1
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