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

Exercise 7.4 · Q13: ∫dx(x−1)(x−2)\int\frac{dx}{\sqrt{(x-1)(x-2)}}

Exercise 7.4Integrals of some particular functions · Section 7.4
The question

Integrate the function (Exercise 7.4, "Integrate the functions in Exercises 1 to 23"):

1(x−1)(x−2)\frac{1}{\sqrt{(x - 1)(x - 2)}}
Exam answer

Let I=∫dx(x−1)(x−2)I = \displaystyle\int\frac{dx}{\sqrt{(x - 1)(x - 2)}}⁠.

(x−1)(x−2)=x2−3x+2=(x−32)2−(12)2\begin{aligned} &(x - 1)(x - 2) \\ &= x^2 - 3x + 2 \\ &= \left(x - \frac32\right)^2 - \left(\frac12\right)^2 \end{aligned}

Put x−32=tx - \frac32 = t⁠. Then dx=dtdx = dt⁠.

I=∫dtt2−(12)2I = \int\frac{dt}{\sqrt{t^2 - \left(\frac12\right)^2}}

[Using ∫dtt2−a2\displaystyle\int\frac{dt}{\sqrt{t^2 - a^2}} =log⁡∣t+t2−a2∣+C= \log\left|t + \sqrt{t^2 - a^2}\right| + C]

I=log⁡∣t+t2−14∣+CI = \log\left|t + \sqrt{t^2 - \frac14}\right| + C

∴ I=log⁡∣x−32+x2−3x+2∣\therefore\ I = \log\left|x - \dfrac32 + \sqrt{x^2 - 3x + 2}\right| +C+ C⁠.

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

The one idea this problem is built on

Multiply out and complete the square: (x−1)(x−2)=x2−3x+2(x - 1)(x - 2) = x^2 - 3x + 2⁠, which equals (x−32)2−(12)2\left(x - \frac32\right)^2 - \left(\frac12\right)^2⁠. Under the root we have "a square minus a constant squared", the shape of formula 7.4 (4): ∫dxx2−a2=log⁡∣x+x2−a2∣+C\int\frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C⁠.

Given, to find, and the plan
Given
f(x)=1(x−1)(x−2)f(x) = \dfrac{1}{\sqrt{(x - 1)(x - 2)}}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Find the domain. (2) Expand and complete the square. (3) Shift t=x−32t = x - \frac32 and use formula 7.4 (4) with a=12a = \frac12⁠. (4) Return to xx⁠. (5) Differentiate back, on both pieces of the domain.

Basics you need

1Sign of a product

(x−1)(x−2)>0(x - 1)(x - 2) > 0 when both factors have the same sign. Basics: quadratics and completing the square

2Completing the square

x2−3x=(x−32)2−94x^2 - 3x = \left(x - \frac32\right)^2 - \frac94⁠. Basics: completing the square

3Formula 7.4 (4)

∫dxx2−a2=log⁡∣x+x2−a2∣+C\int\frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + C⁠, for ∣x∣>a|x| > a⁠. Lesson: integrals of particular functions

Solution

  1. Where is the integrand defined?

    We need (x−1)(x−2)>0(x - 1)(x - 2) > 0⁠. Both factors positive: x>2x > 2⁠. Both negative: x<1x < 1⁠. So the domain is

    x<1orx>2.x < 1 \quad\text{or}\quad x > 2.
    x y 1 2 3 3 6 vertex (1.5, −0.25) y = x² − 3x + 2
    y=(x−1)(x−2)=x2−3x+2y = (x - 1)(x - 2) = x^2 - 3x + 2 is an upward parabola with vertex (32,−14)\left(\frac32, -\frac14\right)⁠. It is positive (green) outside its roots 1 and 2 and negative (rose) between them, so the integrand exists on the two green pieces of the xx-axis.
    The root is in a denominator, so the product must be strictly positive. Between 1 and 2 the factors have opposite signs and the integrand does not exist. The domain has two separate pieces, and the answer must work on both.

  2. Expand and complete the square

    (x−1)(x−2)=x2−3x+2=(x−32)2−94+2=(x−32)2−(12)2.\begin{aligned} (x - 1)(x - 2) &= x^2 - 3x + 2 \\ &= \left(x - \frac32\right)^2 - \frac94 + 2 \\ &= \left(x - \frac32\right)^2 - \left(\frac12\right)^2. \end{aligned}
    Half of −3-3 is −32-\frac32⁠, whose square is 94\frac94⁠. The leftover −94+84=−14=−(12)2-\frac94 + \frac84 = -\frac14 = -\left(\frac12\right)^2 is negative: "square minus square", which leads to formula 7.4 (4).
    Basics: completing the square

  3. Shift t=x−32t = x - \frac32 and use formula 7.4 (4)

    Put t=x−32t = x - \frac32⁠, dt=dxdt = dx⁠. On the domain, t2>14t^2 > \frac14⁠, that is, ∣t∣>12|t| > \frac12⁠:

    ∫dx(x−1)(x−2)=∫dtt2−(12)2=log⁡∣t+t2−14∣+C.\begin{aligned} \int\frac{dx}{\sqrt{(x - 1)(x - 2)}} &= \int\frac{dt}{\sqrt{t^2 - \left(\frac12\right)^2}} \\ &= \log\left|t + \sqrt{t^2 - \tfrac14}\right| + C. \end{aligned}
    Formula 7.4 (4) with a=12a = \frac12⁠. The book proves it with t=asec⁡θt = a\sec\theta⁠, which covers t>at > a (our piece x>2x > 2⁠). For t<−at < -a (our piece x<1x < 1⁠) the same formula is still correct, as the check in Step 5 shows; there t+t2−a2t + \sqrt{t^2 - a^2} is negative, so the modulus is essential.
    Formula 7.4 (4); type (8), Section 7.4. Lesson: integrals of particular functions

  4. Return to xx

    t2−14=(x−1)(x−2)=x2−3x+2t^2 - \frac14 = (x - 1)(x - 2) = x^2 - 3x + 2 (Step 2 read backwards), so

    ∫dx(x−1)(x−2)=log⁡∣x−32+x2−3x+2∣+C.\begin{aligned} &\int\frac{dx}{\sqrt{(x - 1)(x - 2)}} \\ &= \log\left|x - \frac32 + \sqrt{x^2 - 3x + 2}\right| + C. \end{aligned}
    This is the textbook's form. The quantity inside the modulus is never 0, because x2−3x+2<∣x−32∣\sqrt{x^2 - 3x + 2} < \left|x - \frac32\right| (the root is of the square minus 14\frac14⁠).

  5. Check: differentiate the answer back

    Let s=x2−3x+2s = \sqrt{x^2 - 3x + 2}⁠, so s′=2x−32s=x−32ss' = \frac{2x - 3}{2s} = \frac{x - \frac32}{s}⁠. With u=x−32+su = x - \frac32 + s and ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u}⁠,

    F′(x)=1+x−3/2sx−32+s=s+x−32s (x−32+s)=1s.\begin{aligned} F'(x) &= \frac{1 + \frac{x - 3/2}{s}}{x - \frac32 + s} \\ &= \frac{s + x - \frac32}{s\,\left(x - \frac32 + s\right)} = \frac1s. \end{aligned}

    So F′(x)=1(x−1)(x−2)F'(x) = \frac{1}{\sqrt{(x - 1)(x - 2)}}⁠, the integrand. ✓\checkmark

    ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u} holds whether uu is positive or negative, and nothing in the calculation assumed x>2x > 2⁠. So the answer is valid on both pieces, x<1x < 1 and x>2x > 2⁠.
    Standard formula (xii), Section 7.2, with the chain rule. Basics: integration undoes differentiation
Result

∫dx(x−1)(x−2)=log⁡∣x−32+x2−3x+2∣+C,\begin{aligned} &\int\frac{dx}{\sqrt{(x - 1)(x - 2)}} \\ &= \log\left|x - \frac32 + \sqrt{x^2 - 3x + 2}\right| + C, \end{aligned} valid for x<1x < 1 or x>2x > 2⁠. (Textbook answer: log⁡∣x−32+x2−3x+2∣+C\log\left|x - \frac32 + \sqrt{x^2 - 3x + 2}\right| + C⁠, the same.)

Check it with numbers

With h=0.001h = 0.001⁠, once on each piece of the domain:

  • At x=3x = 3⁠: F(3.001)=1.07030684F(3.001) = 1.07030684 and F(2.999)=1.06889262F(2.999) = 1.06889262⁠; slope ≈1.07030684−1.068892620.002=0.7071\approx \frac{1.07030684 - 1.06889262}{0.002} = 0.7071⁠; integrand 12⋅1=0.7071\frac{1}{\sqrt{2\cdot 1}} = 0.7071⁠. ✓
  • At x=0x = 0⁠: F(0.001)=−2.45518698F(0.001) = -2.45518698 and F(−0.001)=−2.45660120F(-0.001) = -2.45660120⁠; slope ≈−2.45518698+2.456601200.002=0.7071\approx \frac{-2.45518698 + 2.45660120}{0.002} = 0.7071⁠; integrand 1(−1)(−2)=0.7071\frac{1}{\sqrt{(-1)(-2)}} = 0.7071⁠. ✓
Common mistakes
  1. Splitting the root as x−1 x−2\sqrt{x - 1}\,\sqrt{x - 2}⁠. That is only allowed when both factors are positive, so it loses the piece x<1x < 1⁠.
  2. Completing the square wrongly: 94−2=14\frac94 - 2 = \frac14⁠, so a=12a = \frac12⁠, not 14\frac14⁠.
  3. Dropping the modulus. For x<1x < 1 the inside of the logarithm is negative.
What to take away
  1. A product of linear factors under a root: expand, complete the square, shift.
  2. Negative leftover with a positive x2x^2 term means formula 7.4 (4); the domain is outside the roots.
  3. Check the answer on every piece of the domain.

Every fact used, and where it comes from

FactUsed inSource
A product is positive when both factors have the same signStep 1Sign rules for real numbers
Completing the squareStep 2Type (8), Section 7.4 (Basics: completing the square)
Shift t=x−32t = x - \frac32⁠, dt=dxdt = dxStep 3Section 7.3.1 (Lesson: substitution)
∫dxx2−a2=log⁡∣x+x2−a2∣+C\int\frac{dx}{\sqrt{x^2 - a^2}} = \log\left|x + \sqrt{x^2 - a^2}\right| + CStep 3Formula 7.4 (4), Section 7.4
ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u}⁠; chain ruleStep 5Standard formula (xii), Section 7.2; Theorem 4, Section 5.3.1
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 →