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Chapter 7 · Integrals

Exercise 7.5 · Q16: ∫dxx(xn+1)\int\frac{dx}{x(x^n + 1)}

Exercise 7.5Substitution and partial fractions · Section 7.5
The question

Integrate the rational function

1x(xn+1)\frac{1}{x(x^n + 1)}

[Hint: multiply numerator and denominator by xn−1x^{n - 1} and put xn=tx^n = t⁠.]

Exam answer

Let I=∫dxx(xn+1)I = \displaystyle\int\frac{dx}{x(x^n + 1)}⁠.

Multiplying numerator and denominator by xn−1x^{n - 1}⁠,

I=∫xn−1 dxxn(xn+1)I = \int\frac{x^{n - 1}\,dx}{x^n(x^n + 1)}

Put xn=tx^n = t⁠. Then nxn−1 dx=dtnx^{n - 1}\,dx = dt⁠.

I=1n∫dtt(t+1)I = \frac1n\int\frac{dt}{t(t + 1)}

Let 1t(t+1)=At+Bt+1\frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1}⁠. Then 1=A(t+1)+Bt1 = A(t + 1) + Bt⁠.

Putting t=0t = 0⁠: A=1A = 1⁠. Putting t=−1t = -1⁠: B=−1B = -1⁠.

∴ I=1n∫(1t−1t+1)dt=1nlog⁡∣t∣−1nlog⁡∣t+1∣+C=1nlog⁡∣tt+1∣+C\begin{aligned} \therefore\ I &= \frac1n\int\left(\frac1t - \frac{1}{t + 1}\right)dt \\ &= \frac1n\log|t| \\ &\quad - \frac1n\log|t + 1| + C \\ &= \frac1n\log\left|\frac{t}{t + 1}\right| + C \end{aligned}

∴ I=1nlog⁡∣xnxn+1∣+C\therefore\ I = \dfrac1n\log\left|\dfrac{x^n}{x^n + 1}\right| + C⁠.

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

The one idea this problem is built on

The derivative of xnx^n is nxn−1nx^{n - 1}⁠. Multiplying top and bottom by xn−1x^{n - 1} creates this derivative in the numerator, so the substitution t=xnt = x^n turns the integral into 1n∫dtt(t+1)\frac1n\int\frac{dt}{t(t + 1)}⁠, which splits into two simple fractions (Table 7.2, row 1).

Given, to find, and the plan
Given
The integrand 1x(xn+1)\frac{1}{x(x^n + 1)}⁠, where nn is a fixed positive integer (the usual reading; the same working holds for any real n≠0n \ne 0 when x>0x > 0⁠). It is defined when x≠0x \ne 0 and xn≠−1x^n \ne -1 (for odd nn this excludes x=−1x = -1⁠; for even nn⁠, xn≥0x^n \ge 0 so nothing more is excluded).
To find
∫dxx(xn+1)\int\frac{dx}{x(x^n + 1)}⁠.
Plan
(1) Multiply by xn−1xn−1\frac{x^{n - 1}}{x^{n - 1}}⁠. (2) Put t=xnt = x^n⁠. (3) Split 1t(t+1)\frac{1}{t(t + 1)}⁠. (4) Integrate and return to xx⁠. (5) Differentiate back and test small nn⁠.

Basics you need

1Derivative of xnx^n

ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n - 1} (standard formula (i), Section 7.2, read as a derivative). Basics: powers of x

2Substitution

If t=g(x)t = g(x) then dt=g′(x) dxdt = g'(x)\,dx⁠, and ∫f(g(x)) g′(x) dx=∫f(t) dt\int f(g(x))\,g'(x)\,dx = \int f(t)\,dt⁠. Basics: substitution

3Table 7.2, row 1

1(t−a)(t−b)=At−a+Bt−b\frac{1}{(t - a)(t - b)} = \frac{A}{t - a} + \frac{B}{t - b}⁠, a≠ba \ne b⁠. Lesson: partial fractions

4Logarithm laws

log⁡∣p∣−log⁡∣q∣=log⁡∣pq∣\log|p| - \log|q| = \log\left|\frac pq\right| and log⁡∣xn∣=nlog⁡∣x∣\log|x^n| = n\log|x|⁠. Basics: logarithms and 1/x

Solution

  1. Multiply numerator and denominator by xn−1x^{n - 1}

    1x(xn+1)=xn−1x⋅xn−1(xn+1)=xn−1xn(xn+1).\frac{1}{x(x^n + 1)} = \frac{x^{n - 1}}{x\cdot x^{n - 1}(x^n + 1)} = \frac{x^{n - 1}}{x^n(x^n + 1)}.
    Since x≠0x \ne 0⁠, xn−1≠0x^{n - 1} \ne 0⁠, so multiplying top and bottom by it does not change the value. The numerator is now 1n\frac1n times the derivative of xnx^n⁠, and the denominator is written entirely in terms of xnx^n⁠. This prepares the substitution.
    x⋅xn−1=xnx\cdot x^{n - 1} = x^n (law of indices). Basics: powers of x

  2. Substitute t=xnt = x^n

    Then dt=nxn−1 dxdt = nx^{n - 1}\,dx⁠, so xn−1 dx=1n dtx^{n - 1}\,dx = \frac1n\,dt⁠. Hence

    ∫xn−1 dxxn(xn+1)=1n∫dtt(t+1).\int\frac{x^{n - 1}\,dx}{x^n(x^n + 1)} = \frac1n\int\frac{dt}{t(t + 1)}.

    Here t≠0t \ne 0 (as x≠0x \ne 0⁠) and t≠−1t \ne -1 (as xn≠−1x^n \ne -1⁠).

    The substitution swaps the awkward power xnx^n for a single letter. The factor xn−1dxx^{n - 1}dx is exactly what dtdt supplies, up to the constant 1n\frac1n (which needs n≠0n \ne 0⁠).
    Integration by substitution, Section 7.3.1. Lesson: substitution

  3. Split 1t(t+1)\frac{1}{t(t + 1)} into partial fractions

    1t(t+1)=At+Bt+1,1=A(t+1)+Bt.\begin{gathered} \frac{1}{t(t + 1)} = \frac{A}{t} + \frac{B}{t + 1}, \\ 1 = A(t + 1) + Bt. \end{gathered}

    Put t=0t = 0⁠: A=1A = 1⁠. Put t=−1t = -1⁠: 1=−B1 = -B⁠, B=−1B = -1⁠. So

    1t(t+1)=1t−1t+1.\frac{1}{t(t + 1)} = \frac1t - \frac{1}{t + 1}.
    The denominator has two different linear factors, t−0t - 0 and t−(−1)t - (-1)⁠: row 1 of Table 7.2. The cleared identity holds for all tt⁠, so the roots may be substituted. (Comparing coefficients gives the same: A+B=0A + B = 0⁠, A=1A = 1⁠.)
    Table 7.2, row 1. Basics: splitting fractions

  4. Integrate and return to xx

    1n∫(1t−1t+1)dt=1n(log⁡∣t∣−log⁡∣t+1∣)+C=1nlog⁡∣tt+1∣+C=1nlog⁡∣xnxn+1∣+C.\begin{aligned} \frac1n\int\Big(\frac1t - \frac{1}{t + 1}\Big)dt &= \frac1n\big(\log|t| - \log|t + 1|\big) + C \\ &= \frac1n\log\left|\frac{t}{t + 1}\right| + C \\ &= \frac1n\log\left|\frac{x^n}{x^n + 1}\right| + C. \end{aligned}
    Each simple fraction integrates to a logarithm (standard formula (xii)); the difference of two logs is the log of the quotient. Finally tt is replaced by xnx^n⁠, because the question is about xx⁠.
    Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

  5. Check: differentiate the answer

    Write the answer as 1n(log⁡∣xn∣−log⁡∣xn+1∣)\frac1n\big(\log|x^n| - \log|x^n + 1|\big)⁠, that is, log⁡∣x∣−1nlog⁡∣xn+1∣\log|x| - \frac1n\log|x^n + 1|⁠. Then

    ddx(log⁡∣x∣−1nlog⁡∣xn+1∣)=1x−1n⋅nxn−1xn+1=(xn+1)−x⋅xn−1x(xn+1)=1x(xn+1). ✓\begin{aligned} &\frac{d}{dx}\Big(\log|x| - \frac1n\log|x^n + 1|\Big) \\ &= \frac1x - \frac1n\cdot\frac{nx^{n - 1}}{x^n + 1} \\ &= \frac{(x^n + 1) - x\cdot x^{n - 1}}{x(x^n + 1)} \\ &= \frac{1}{x(x^n + 1)}. \ \checkmark \end{aligned}
    log⁡∣xn∣=nlog⁡∣x∣\log|x^n| = n\log|x|⁠, so 1nlog⁡∣xn∣=log⁡∣x∣\frac1n\log|x^n| = \log|x|⁠. The chain rule on log⁡∣xn+1∣\log|x^n + 1| gives nxn−1xn+1\frac{nx^{n - 1}}{x^n + 1}⁠. Over the common denominator, x⋅xn−1=xnx\cdot x^{n - 1} = x^n cancels, leaving 1. This also shows a second route: 1x(xn+1)=1x−xn−1xn+1\frac{1}{x(x^n + 1)} = \frac1x - \frac{x^{n - 1}}{x^n + 1} directly.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation

  6. Test small values of nn

    For n=1n = 1 the formula gives ∫dxx(x+1)=log⁡∣xx+1∣+C\int\frac{dx}{x(x + 1)} = \log\left|\frac{x}{x + 1}\right| + C⁠, the familiar row 1 result. For n=2n = 2 it gives 12log⁡x2x2+1+C=log⁡∣x∣−12log⁡(x2+1)+C\frac12\log\frac{x^2}{x^2 + 1} + C = \log|x| - \frac12\log(x^2 + 1) + C⁠, which is exactly option (A) of Exercise 7.5, Q23.

    A general formula must agree with the cases we can do directly. Both agree.
Result ∫dxx(xn+1)=1nlog⁡∣xnxn+1∣+C,\int\frac{dx}{x(x^n + 1)} = \frac1n\log\left|\frac{x^n}{x^n + 1}\right| + C,

where x≠0x \ne 0 and xn≠−1x^n \ne -1⁠. (Textbook answer: 1nlog⁡∣xnxn+1∣+C\frac1n\log\left|\frac{x^n}{x^n + 1}\right| + C⁠, the same.)

Check it with numbers

Take n=3n = 3 and integrate from 1 to 2 (continuous there): F(2)−F(1)F(2) - F(1) =13log⁡89−13log⁡12= \frac13\log\frac89 - \frac13\log\frac12 =13log⁡169≈0.1917880= \frac13\log\frac{16}{9} \approx 0.1917880⁠. A numerical integration of 1x(x3+1)\frac{1}{x(x^3 + 1)} over [1,2][1, 2] gives 0.19178800.1917880 too. ✓

Decomposition at t=1t = 1⁠: 11⋅2=0.5\frac{1}{1\cdot 2} = 0.5 and 11−12=0.5\frac11 - \frac12 = 0.5⁠. ✓

Common mistakes
  1. Forgetting the 1n\frac1n⁠. dt=nxn−1dxdt = nx^{n - 1}dx⁠, so xn−1dx=1ndtx^{n - 1}dx = \frac1n dt⁠, not dtdt⁠.
  2. Leaving tt in the answer. The final answer must be in terms of xx⁠.
  3. Writing log⁡∣xn+1∣−log⁡∣xn∣\log|x^n + 1| - \log|x^n|⁠. That reverses the sign; the 1t\frac1t term is positive, so xnx^n goes on top.
What to take away
  1. Multiplying top and bottom by a well-chosen factor can create the derivative needed for a substitution.
  2. After a substitution, partial fractions work in the new variable exactly as in xx⁠.
  3. 1t(t+1)=1t−1t+1\frac{1}{t(t + 1)} = \frac1t - \frac{1}{t + 1} is worth remembering.

Every fact used, and where it comes from

FactUsed inSource
x⋅xn−1=xnx\cdot x^{n - 1} = x^n⁠; ddxxn=nxn−1\frac{d}{dx}x^n = nx^{n - 1}Steps 1, 2, 5Laws of indices; standard formula (i), Section 7.2
Substitution t=xnt = x^nStep 2Section 7.3.1 (Lesson: substitution)
At−a+Bt−b\frac{A}{t - a} + \frac{B}{t - b}Step 3Table 7.2, row 1, Section 7.5
∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 4Standard formula (xii), Section 7.2
log⁡∣p∣−log⁡∣q∣=log⁡∣pq∣\log|p| - \log|q| = \log\left|\frac pq\right|⁠; log⁡∣xn∣=nlog⁡∣x∣\log|x^n| = n\log|x|Steps 4, 5Laws of logarithms (Basics: logarithms)
Derivative of an antiderivative is the integrandStep 5Property (I), Section 7.2.1
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