Class 12 Maths, Worked in Full Past papersGet updates
Chapter 7 · Integrals

Exercise 7.5 · Q19: ∫2x(x2+1)(x2+3) dx\int\frac{2x}{(x^2 + 1)(x^2 + 3)}\,dx

Exercise 7.5Substitution and partial fractions · Section 7.5
The question

Integrate the rational function

2x(x2+1)(x2+3).\frac{2x}{(x^2 + 1)(x^2 + 3)}.
Exam answer

Let I=∫2x dx(x2+1)(x2+3)I = \displaystyle\int\frac{2x\,dx}{(x^2 + 1)(x^2 + 3)}⁠.

Put x2=tx^2 = t⁠. Then 2x dx=dt2x\,dx = dt⁠.

I=∫dt(t+1)(t+3)I = \int\frac{dt}{(t + 1)(t + 3)}

Let

1(t+1)(t+3)=At+1+Bt+3\frac{1}{(t + 1)(t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3} ⇒ 1=A(t+3)+B(t+1)\Rightarrow\ 1 = A(t + 3) + B(t + 1)

Putting t=−1t = -1⁠: A=12A = \frac12⁠. Putting t=−3t = -3⁠: B=−12B = -\frac12⁠.

∴ I=12∫dtt+1−12∫dtt+3=12log⁡∣t+1∣−12log⁡∣t+3∣+C=12log⁡∣t+1t+3∣+C\begin{aligned} \therefore\ I &= \frac12\int\frac{dt}{t + 1} - \frac12\int\frac{dt}{t + 3} \\ &= \frac12\log|t + 1| \\ &\quad - \frac12\log|t + 3| + C \\ &= \frac12\log\left|\frac{t + 1}{t + 3}\right| + C \end{aligned}

∴ I=12log⁡(x2+1x2+3)+C\therefore\ I = \dfrac12\log\left(\dfrac{x^2 + 1}{x^2 + 3}\right) + C⁠.

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

The one idea this problem is built on

The numerator 2x2x is the derivative of x2x^2⁠, and the denominator is built only from x2x^2⁠. So t=x2t = x^2 gives ∫dt(t+1)(t+3)\int\frac{dt}{(t + 1)(t + 3)}⁠: two different linear factors in tt⁠, split by Table 7.2 (row 1) into two logarithms.

Given, to find, and the plan
Given
The integrand 2x(x2+1)(x2+3)\frac{2x}{(x^2 + 1)(x^2 + 3)}⁠, defined for every real xx (both factors are positive).
To find
∫2x dx(x2+1)(x2+3)\int\frac{2x\,dx}{(x^2 + 1)(x^2 + 3)}⁠.
Plan
(1) Put t=x2t = x^2⁠. (2) Split 1(t+1)(t+3)\frac{1}{(t + 1)(t + 3)}⁠. (3) Integrate and return to xx⁠. (4) Differentiate back.

Basics you need

1Substitution

With t=x2t = x^2⁠, dt=2x dxdt = 2x\,dx⁠. Basics: substitution

2Table 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

3Logarithms

∫dtt+k=log⁡∣t+k∣+C\int\frac{dt}{t + k} = \log|t + k| + C⁠; log⁡p−log⁡q=log⁡pq\log p - \log q = \log\frac pq⁠. Basics: logarithms and 1/x

Solution

  1. Substitute t=x2t = x^2

    Then dt=2x dxdt = 2x\,dx⁠, and

    ∫2x dx(x2+1)(x2+3)=∫dt(t+1)(t+3),\int\frac{2x\,dx}{(x^2 + 1)(x^2 + 3)} = \int\frac{dt}{(t + 1)(t + 3)},

    with t≥0t \ge 0⁠.

    The product 2x dx2x\,dx is exactly dtdt⁠, and the rest of the integrand depends on xx only through x2=tx^2 = t⁠. (Here, unlike Example 14, t=x2t = x^2 is a genuine change of variable in the integral, because dtdt is present.)
    Integration by substitution, Section 7.3.1. Lesson: substitution

  2. Split into partial fractions

    1(t+1)(t+3)=At+1+Bt+3,1=A(t+3)+B(t+1).\begin{gathered} \frac{1}{(t + 1)(t + 3)} = \frac{A}{t + 1} + \frac{B}{t + 3}, \\ 1 = A(t + 3) + B(t + 1). \end{gathered}

    Put t=−1t = -1⁠: 1=2A1 = 2A⁠, A=12A = \frac12⁠. Put t=−3t = -3⁠: 1=−2B1 = -2B⁠, B=−12B = -\frac12⁠. So

    1(t+1)(t+3)=12(1t+1−1t+3).\frac{1}{(t + 1)(t + 3)} = \frac12\Big(\frac{1}{t + 1} - \frac{1}{t + 3}\Big).
    Two different linear factors: row 1 of Table 7.2. The cleared identity is a polynomial identity valid for every tt⁠, so the roots may be substituted even though t=x2t = x^2 never takes these values.
    Table 7.2, row 1. Basics: splitting fractions

  3. Integrate and return to xx

    12∫(1t+1−1t+3)dt=12log⁡∣t+1t+3∣+C=12log⁡x2+1x2+3+C.\begin{aligned} \frac12\int\Big(\frac{1}{t + 1} - \frac{1}{t + 3}\Big)dt &= \frac12\log\left|\frac{t + 1}{t + 3}\right| + C \\ &= \frac12\log\frac{x^2 + 1}{x^2 + 3} + C. \end{aligned}
    Each simple fraction gives a logarithm (standard formula (xii)), and the difference of logs is the log of the quotient. Since x2+1>0x^2 + 1 > 0 and x2+3>0x^2 + 3 > 0⁠, the quotient is positive and the modulus can be dropped.
    Standard formula (xii), Section 7.2. Basics: logarithms and 1/x

  4. Check: differentiate the answer

    Chain rule on each logarithm (inner derivatives 2x2x⁠):

    ddx(12log⁡(x2+1)−12log⁡(x2+3))=xx2+1−xx2+3=x[(x2+3)−(x2+1)](x2+1)(x2+3)=2x(x2+1)(x2+3). ✓\begin{aligned} &\frac{d}{dx}\Big(\tfrac12\log(x^2 + 1) - \tfrac12\log(x^2 + 3)\Big) \\ &= \frac{x}{x^2 + 1} - \frac{x}{x^2 + 3} \\ &= \frac{x\big[(x^2 + 3) - (x^2 + 1)\big]}{(x^2 + 1)(x^2 + 3)} \\ &= \frac{2x}{(x^2 + 1)(x^2 + 3)}. \ \checkmark \end{aligned}

    Valid for every real xx⁠.

    12⋅2xx2+1=xx2+1\frac12\cdot\frac{2x}{x^2 + 1} = \frac{x}{x^2 + 1}⁠. Over the common denominator the bracket is 3−1=23 - 1 = 2⁠, giving the original numerator 2x2x⁠.
    Property (I), Section 7.2.1. Basics: integration undoes differentiation
Result ∫2x dx(x2+1)(x2+3)=12log⁡(x2+1x2+3)+C,\int\frac{2x\,dx}{(x^2 + 1)(x^2 + 3)} = \frac12\log\Big(\frac{x^2 + 1}{x^2 + 3}\Big) + C,

for every real xx⁠. (Textbook answer: 12log⁡(x2+1x2+3)+C\frac12\log\big(\frac{x^2 + 1}{x^2 + 3}\big) + C⁠, the same.)

Check it with numbers

Decomposition at t=1t = 1⁠: 12⋅4=0.125\frac{1}{2\cdot 4} = 0.125 and 12(12−14)=0.125\frac12\big(\frac12 - \frac14\big) = 0.125⁠. ✓

Definite integral from 0 to 1: F(1)−F(0)=12log⁡24−12log⁡13F(1) - F(0) = \frac12\log\frac24 - \frac12\log\frac13 =12log⁡32≈0.2027326= \frac12\log\frac32 \approx 0.2027326⁠. A numerical integration over [0,1][0, 1] gives 0.20273260.2027326 too. ✓

Common mistakes
  1. Losing the 12\frac12⁠. A=12A = \frac12 and B=−12B = -\frac12⁠, not 11 and −1-1⁠: check by putting t=0t = 0⁠, where 13\frac13 must equal A+B3A + \frac B3⁠.
  2. Treating the fraction as needing Bx+CBx + C numerators. After t=x2t = x^2 the factors are linear in tt⁠, so constants suffice.
  3. Leaving the answer in tt⁠. Replace tt by x2x^2⁠.
What to take away
  1. Spot the derivative of the inner quantity in the numerator (2x2x for x2x^2⁠) and substitute.
  2. 1(t+a)(t+b)=1b−a(1t+a−1t+b)\frac{1}{(t + a)(t + b)} = \frac{1}{b - a}\Big(\frac{1}{t + a} - \frac{1}{t + b}\Big) for a≠ba \ne b⁠.
  3. Drop the modulus only when the quantity is known to be positive.

Every fact used, and where it comes from

FactUsed inSource
Substitution t=x2t = x^2⁠, dt=2x dxdt = 2x\,dxStep 1Section 7.3.1 (Lesson: substitution)
At−a+Bt−b\frac{A}{t - a} + \frac{B}{t - b}Step 2Table 7.2, row 1, Section 7.5
∫dtt=log⁡∣t∣+C\int\frac{dt}{t} = \log|t| + CStep 3Standard formula (xii), Section 7.2
log⁡p−log⁡q=log⁡pq\log p - \log q = \log\frac pqStep 3Laws of logarithms (Basics: logarithms)
Derivative of an antiderivative is the integrandStep 4Property (I), Section 7.2.1
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →