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

Exercise 7.4 · Q7: ∫x−1x2−1 dx\int\frac{x-1}{\sqrt{x^2-1}}\,dx

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"):

x−1x2−1\frac{x - 1}{\sqrt{x^2 - 1}}
Exam answer

Let I=∫x−1x2−1 dxI = \displaystyle\int\frac{x - 1}{\sqrt{x^2 - 1}}\,dx⁠.

I=∫x dxx2−1−∫dxx2−1=I1−I2\begin{aligned} I &= \int\frac{x\,dx}{\sqrt{x^2 - 1}} - \int\frac{dx}{\sqrt{x^2 - 1}} \\ &= I_1 - I_2 \end{aligned}

In I1I_1⁠, put x2−1=tx^2 - 1 = t⁠. Then 2x dx=dt2x\,dx = dt⁠.

I1=12∫dtt=t=x2−1\begin{aligned} I_1 &= \frac12\int\frac{dt}{\sqrt t} = \sqrt t \\ &= \sqrt{x^2 - 1} \end{aligned}

Using ∫dxx2−a2\displaystyle\int\frac{dx}{\sqrt{x^2 - a^2}} =log⁡∣x+x2−a2∣+C= \log\left|x + \sqrt{x^2 - a^2}\right| + C⁠,

I2=log⁡∣x+x2−1∣I_2 = \log\left|x + \sqrt{x^2 - 1}\right|

∴ I=x2−1−log⁡∣x+x2−1∣\therefore\ I = \sqrt{x^2 - 1} - \log\left|x + \sqrt{x^2 - 1}\right| +C+ C⁠.

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

The one idea this problem is built on

Split the numerator: x−1x2−1=xx2−1−1x2−1\frac{x - 1}{\sqrt{x^2 - 1}} = \frac{x}{\sqrt{x^2 - 1}} - \frac{1}{\sqrt{x^2 - 1}}⁠. The first piece has the derivative of x2−1x^2 - 1 (up to a factor 2) on top, so a substitution gives a square root. The second piece is the standard formula 7.4 (4), a logarithm.

Given, to find, and the plan
Given
f(x)=x−1x2−1f(x) = \dfrac{x - 1}{\sqrt{x^2 - 1}}⁠.
To find
∫f(x) dx\int f(x)\,dx⁠, and where the answer is valid.
Plan
(1) Find where ff is defined. (2) Split into two integrals. (3) First integral by t=x2−1t = x^2 - 1⁠. (4) Second by formula 7.4 (4). (5) Combine. (6) Differentiate back on both pieces of the domain.

Basics you need

1Splitting a fraction

p−qr=pr−qr\frac{p - q}{r} = \frac pr - \frac qr⁠, and ∫(f−g)=∫f−∫g\int(f - g) = \int f - \int g (Properties (III), (IV), Section 7.2.1). Basics: splitting fractions

2Power rule after substitution

∫t−1/2 dt=t1/21/2=2t+C\int t^{-1/2}\,dt = \frac{t^{1/2}}{1/2} = 2\sqrt t + C⁠. Basics: the power rule

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. Lesson: integrals of particular functions

4Substitution

t=x2−1t = x^2 - 1 gives dt=2x dxdt = 2x\,dx⁠. Basics: substitution (the chain rule backwards)

Solution

  1. Where is the integrand defined?

    We need x2−1>0x^2 - 1 > 0⁠, that is, ∣x∣>1|x| > 1⁠: either x>1x > 1 or x<−1x < -1⁠.

    The root is in the denominator, so the quantity under it must be strictly positive. The domain has two separate pieces, and the answer must be checked on both.

  2. Split into two integrals

    ∫x−1x2−1 dx=∫x dxx2−1−∫dxx2−1.\int\frac{x - 1}{\sqrt{x^2 - 1}}\,dx = \int\frac{x\,dx}{\sqrt{x^2 - 1}} - \int\frac{dx}{\sqrt{x^2 - 1}}.

    Call these I1I_1 and I2I_2⁠.

    Neither standard form fits the whole fraction, but each piece fits one. The xx on top of I1I_1 is half the derivative of x2−1x^2 - 1⁠; I2I_2 is formula 7.4 (4) exactly. This split is a simple case of type (10) in Section 7.4, where px+qpx + q is written as Addx(quadratic)+BA\frac{d}{dx}(\text{quadratic}) + B⁠: here x−1=12(2x)−1x - 1 = \frac12(2x) - 1⁠.
    Properties (III) and (IV), Section 7.2.1. Basics: splitting fractions

  3. First integral: substitute t=x2−1t = x^2 - 1

    Put t=x2−1t = x^2 - 1⁠, so dt=2x dxdt = 2x\,dx⁠, x dx=12 dtx\,dx = \frac12\,dt⁠, with t>0t > 0 on the domain:

    I1=12∫t−1/2 dt=12⋅2t1/2+C1=x2−1+C1.\begin{aligned} I_1 &= \frac12\int t^{-1/2}\,dt = \frac12\cdot 2t^{1/2} + C_1 \\ &= \sqrt{x^2 - 1} + C_1. \end{aligned}
    After the substitution the integrand is a plain power of tt⁠, so the power rule applies with n=−12n = -\frac12⁠. The substitution works on both pieces of the domain, because t=x2−1>0t = x^2 - 1 > 0 whether x>1x > 1 or x<−1x < -1⁠.
    Power rule, Section 7.2; substitution, Section 7.3.1. Lesson: integration by substitution

  4. Second integral: formula 7.4 (4) with a=1a = 1

    I2=∫dxx2−12=log⁡∣x+x2−1∣+C2.I_2 = \int\frac{dx}{\sqrt{x^2 - 1^2}} = \log\left|x + \sqrt{x^2 - 1}\right| + C_2.
    "Square minus constant, under a root" is formula 7.4 (4). The book proves it with x=asec⁡θx = a\sec\theta⁠, which handles x>ax > a⁠; the same answer is valid for x<−ax < -a as well, as the check in Step 6 shows. On x<−1x < -1 the quantity x+x2−1x + \sqrt{x^2 - 1} is negative (because x2−1<∣x∣\sqrt{x^2 - 1} < |x|⁠), so the modulus is essential there.
    Formula 7.4 (4). Lesson: integrals of particular functions

  5. Combine

    ∫x−1x2−1 dx=I1−I2=x2−1−log⁡∣x+x2−1∣+C,\begin{aligned} &\int\frac{x - 1}{\sqrt{x^2 - 1}}\,dx = I_1 - I_2 \\ &= \sqrt{x^2 - 1} - \log\left|x + \sqrt{x^2 - 1}\right| + C, \end{aligned}

    where C=C1−C2C = C_1 - C_2⁠.

    Two arbitrary constants combine into one arbitrary constant.

  6. Check: differentiate the answer back

    Let s=x2−1s = \sqrt{x^2 - 1}⁠, so s′=2x2s=xss' = \frac{2x}{2s} = \frac xs⁠. For the logarithm, ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u} with u=x+su = x + s⁠:

    ddxlog⁡∣x+s∣=1+xsx+s=s+xs (x+s)=1s.\frac{d}{dx}\log|x + s| = \frac{1 + \frac xs}{x + s} = \frac{s + x}{s\,(x + s)} = \frac1s.

    So

    F′(x)=xs−1s=x−1x2−1. ✓F'(x) = \frac xs - \frac1s = \frac{x - 1}{\sqrt{x^2 - 1}}. \ \checkmark
    The formula 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>1x > 1⁠. So the answer is correct on both pieces x>1x > 1 and x<−1x < -1⁠.
    Standard formula (xii), Section 7.2, with the chain rule. Basics: integration undoes differentiation
Result

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

Check it with numbers

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

  • At x=2x = 2⁠: F(2.001)=0.41567036F(2.001) = 0.41567036 and F(1.999)=0.41451566F(1.999) = 0.41451566⁠; slope ≈0.41567036−0.414515660.002=0.5774\approx \frac{0.41567036 - 0.41451566}{0.002} = 0.5774⁠; integrand 13=0.5774\frac{1}{\sqrt3} = 0.5774⁠. ✓
  • At x=−2x = -2⁠: F(−1.999)=3.04727636F(-1.999) = 3.04727636 and F(−2.001)=3.05074047F(-2.001) = 3.05074047⁠; slope ≈3.04727636−3.050740470.002=−1.7321\approx \frac{3.04727636 - 3.05074047}{0.002} = -1.7321⁠; integrand −33=−3=−1.7321\frac{-3}{\sqrt3} = -\sqrt3 = -1.7321⁠. ✓
Common mistakes
  1. Cancelling x−1x - 1 against x2−1\sqrt{x^2 - 1} carelessly. x−1x2−1=x−1x+1\frac{x - 1}{\sqrt{x^2 - 1}} = \sqrt{\frac{x - 1}{x + 1}} only for x>1x > 1⁠; for x<−1x < -1 the left side is negative. Splitting avoids this trap.
  2. Forgetting the 12\frac12 in I1I_1, which would give 2x2−12\sqrt{x^2 - 1}⁠.
  3. Dropping the modulus in log⁡∣x+x2−1∣\log\left|x + \sqrt{x^2 - 1}\right|⁠: on x<−1x < -1 the inside is negative.
What to take away
  1. For px+qquadratic\frac{px + q}{\sqrt{\text{quadratic}}}⁠, split into "(derivative of the quadratic) over root" plus "constant over root".
  2. ∫x dxx2−a2=x2−a2+C\int\frac{x\,dx}{\sqrt{x^2 - a^2}} = \sqrt{x^2 - a^2} + C and ∫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.
  3. When the domain has two pieces, check the answer on both.

Every fact used, and where it comes from

FactUsed inSource
∫(f−g)=∫f−∫g\int(f - g) = \int f - \int g⁠; constants come outStep 2Properties (III), (IV), Section 7.2.1
Writing px+qpx + q as Addx(quadratic)+BA\frac{d}{dx}(\text{quadratic}) + BStep 2Type (10), Section 7.4
Substitution t=x2−1t = x^2 - 1⁠; power ruleStep 3Section 7.3.1; standard formula (i), Section 7.2
∫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 4Formula 7.4 (4), Section 7.4
ddxlog⁡∣u∣=u′u\frac{d}{dx}\log|u| = \frac{u'}{u}⁠; chain ruleStep 6Standard 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 →