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

Exercise 7.8 · Q19: ∫026x+3x2+4 dx\int_0^2 \frac{6x + 3}{x^2 + 4}\,dx

Exercise 7.8Fundamental theorem of calculus · Section 7.8
The question

Evaluate the definite integral

∫026x+3x2+4 dx.\int_0^2 \frac{6x + 3}{x^2 + 4}\,dx.
Exam answer

Let I=∫026x+3x2+4 dxI = \displaystyle\int_0^2 \frac{6x + 3}{x^2 + 4}\,dx⁠.

I=3∫022x dxx2+4+3∫02dxx2+22=3[log⁡(x2+4)]02+3⋅12[tan⁡−1x2]02=3(log⁡8−log⁡4)+32(tan⁡−11−tan⁡−10)=3log⁡2+32⋅π4\begin{aligned} I &= 3\int_0^2\frac{2x\,dx}{x^2 + 4} + 3\int_0^2\frac{dx}{x^2 + 2^2} \\ &= 3\Big[\log(x^2 + 4)\Big]_0^2 \\ &\quad + 3\cdot\frac12\Big[\tan^{-1}\frac{x}{2}\Big]_0^2 \\ &= 3(\log 8 - \log 4) \\ &\quad + \frac32\left(\tan^{-1}1 - \tan^{-1}0\right) \\ &= 3\log 2 + \frac32\cdot\frac{\pi}{4} \end{aligned}

∴ I=3log⁡2+3π8\therefore\ I = 3\log 2 + \dfrac{3\pi}{8}⁠.

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

The one idea this problem is built on

Split the numerator. 6xx2+4\frac{6x}{x^2 + 4} has 3 times the derivative of the denominator on top, so it integrates to 3log⁡(x2+4)3\log(x^2 + 4)⁠. 3x2+4\frac{3}{x^2 + 4} is a constant over x2+22x^2 + 2^2⁠, so formula 7.4 (3) gives 32tan⁡−1x2\frac32\tan^{-1}\frac{x}{2}⁠. Then Theorem 2 (Section 7.8.3) gives F(2)−F(0)F(2) - F(0)⁠.

Given, to find, and the plan
Given
The integrand f(x)=6x+3x2+4f(x) = \frac{6x + 3}{x^2 + 4} on [0,2][0, 2]⁠.
To find
The exact value of the integral, and a decimal.
Plan
(1) Check continuity. (2) Split the fraction. (3) Integrate each piece. (4) Differentiate FF back. (5) Compute F(2)−F(0)F(2) - F(0)⁠.

Basics you need

1Second fundamental theorem

∫abf(x) dx=F(b)−F(a)\int_a^b f(x)\,dx = F(b) - F(a) when ff is continuous on [a,b][a, b] and F′=fF' = f⁠. Lesson: definite integrals and the fundamental theorem

2Substitution

t=g(x)t = g(x)⁠, dt=g′(x) dxdt = g'(x)\,dx⁠. Basics: substitution (the chain rule backwards)

3Formula 7.4 (3)

∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + C for a>0a > 0⁠. Lesson: integrals of particular functions

4Logs and inverse tangent

log⁡8−log⁡4=log⁡2\log 8 - \log 4 = \log 2⁠; tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} because π4\frac{\pi}{4} lies in the principal range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠. Basics: logarithms and 1/x

Solution

  1. Check continuity on [0,2][0, 2]

    Numerator and denominator are polynomials, and x2+4≥4>0x^2 + 4 \ge 4 > 0 for every real xx⁠. So the integrand is continuous on all of R\mathbb{R}⁠, in particular on [0,2][0, 2]⁠.

    Theorem 2 needs continuity on the closed interval; the denominator can never be 0.
    x2≥0x^2 \ge 0⁠, so x2+4≥4x^2 + 4 \ge 4⁠. Lesson: the fundamental theorem

  2. Split the fraction

    6x+3x2+4=6xx2+4+3x2+4.\frac{6x + 3}{x^2 + 4} = \frac{6x}{x^2 + 4} + \frac{3}{x^2 + 4}.
    The derivative of x2+4x^2 + 4 is 2x2x⁠, and 6x=3(2x)6x = 3(2x)⁠: the first piece is a log. The second piece has only a constant on top: an inverse tangent.
    p+qr=pr+qr\frac{p + q}{r} = \frac{p}{r} + \frac{q}{r}⁠. Basics: splitting fractions

  3. Integrate each piece

    First piece: put t=x2+4t = x^2 + 4⁠, dt=2x dxdt = 2x\,dx⁠, so 6x dx=3 dt6x\,dx = 3\,dt⁠:

    ∫6x dxx2+4=3∫dtt=3log⁡(x2+4).\int\frac{6x\,dx}{x^2 + 4} = 3\int\frac{dt}{t} = 3\log(x^2 + 4).

    Second piece: formula 7.4 (3) with a=2a = 2⁠:

    ∫3 dxx2+22=3⋅12tan⁡−1x2=32tan⁡−1x2.\int\frac{3\,dx}{x^2 + 2^2} = 3\cdot\frac12\tan^{-1}\frac{x}{2} = \frac32\tan^{-1}\frac{x}{2}.

    So F(x)=3log⁡(x2+4)+32tan⁡−1x2F(x) = 3\log(x^2 + 4) + \frac32\tan^{-1}\frac{x}{2}⁠.

    log⁡∣t∣=log⁡t\log|t| = \log t here because t=x2+4>0t = x^2 + 4 > 0⁠. In formula 7.4 (3), a2=4a^2 = 4 gives a=2a = 2⁠, so the factor is 1a=12\frac1a = \frac12⁠.
    Standard formula (xii), Section 7.2; formula (3), Section 7.4. Lesson: integrals of particular functions

  4. Check: differentiate FF back

    F′(x)=3⋅2xx2+4+32⋅11+x24⋅12=6xx2+4+34⋅4x2+4=6x+3x2+4. ✓\begin{aligned}F'(x) &= 3\cdot\frac{2x}{x^2 + 4} + \frac32\cdot\frac{1}{1 + \frac{x^2}{4}}\cdot\frac12 \\ &= \frac{6x}{x^2 + 4} + \frac34\cdot\frac{4}{x^2 + 4} \\ &= \frac{6x + 3}{x^2 + 4}.\ \checkmark\end{aligned}
    For the inverse tangent, the chain rule with inner function u=x2u = \frac{x}{2} gives the factor 12\frac12⁠, and 11+x24=44+x2\frac{1}{1 + \frac{x^2}{4}} = \frac{4}{4 + x^2}⁠.
    Standard formula (x), Section 7.2: ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}⁠. Basics: integration undoes differentiation

  5. Evaluate F(2)−F(0)F(2) - F(0)

    At x=2x = 2⁠: x2+4=8x^2 + 4 = 8 and tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠. At x=0x = 0⁠: x2+4=4x^2 + 4 = 4 and tan⁡−10=0\tan^{-1}0 = 0⁠. So

    I=[3log⁡8+32⋅π4]−[3log⁡4+0]=3log⁡84+3π8=3log⁡2+3π8.\begin{aligned}I &= \left[3\log 8 + \tfrac32\cdot\tfrac{\pi}{4}\right] - \left[3\log 4 + 0\right] \\ &= 3\log\frac84 + \frac{3\pi}{8} = 3\log 2 + \frac{3\pi}{8}.\end{aligned}

    In decimals: 3log⁡2≈2.079441543\log 2 \approx 2.07944154 and 3π8≈1.17809725\frac{3\pi}{8} \approx 1.17809725⁠, so I≈3.25753879I \approx 3.25753879⁠, that is I≈3.2575388I \approx 3.2575388⁠.

    1 2 3 0.5 1 1.5 2 x y y = (6x + 3)/(x² + 4) area ≈ 3.258
    The shaded region under y=6x+3x2+4y = \frac{6x + 3}{x^2 + 4} from 0 to 2 has area about 3.258. The curve rises from 0.75 to a peak of about 1.92 near x=1.56x = 1.56 and is 1.875 at x=2x = 2⁠.
    Note log⁡4≠0\log 4 \ne 0⁠: the lower limit contributes 3log⁡43\log 4⁠, which combines with 3log⁡83\log 8 into 3log⁡23\log 2⁠. The figure is a sanity check: heights between 0.75 and 1.92 over a width of 2 give an area between 1.5 and 3.84.
    tan⁡π4=1\tan\frac{\pi}{4} = 1⁠, so tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} (principal value). Basics: trigonometric values
Result ∫026x+3x2+4 dx=3log⁡2+3π8≈3.2575388.\int_0^2\frac{6x + 3}{x^2 + 4}\,dx = 3\log 2 + \frac{3\pi}{8} \approx 3.2575388.

(Textbook answer: 3log⁡2+3π83\log 2 + \frac{3\pi}{8}⁠.)

Check it with numbers

A numerical integration of 6x+3x2+4\frac{6x + 3}{x^2 + 4} over [0,2][0, 2] gives about 3.2575388, the same as the exact value to 7 decimal places.

Common mistakes
  1. Writing 3tan⁡−1x23\tan^{-1}\frac{x}{2}⁠. Formula 7.4 (3) carries the factor 1a=12\frac1a = \frac12⁠.
  2. Dropping 3log⁡43\log 4 at the lower limit. log⁡(02+4)=log⁡4\log(0^2 + 4) = \log 4⁠, not 0.
  3. Using degrees. tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4} radians, not 45 as a number.
What to take away
  1. px+qx2+a2\frac{px + q}{x^2 + a^2}⁠: the pxpx part gives p2log⁡(x2+a2)\frac{p}{2}\log(x^2 + a^2)⁠, the qq part gives qatan⁡−1xa\frac{q}{a}\tan^{-1}\frac{x}{a}⁠.
  2. Evaluate every term at both limits; a log does not vanish at x=0x = 0 unless its argument is 1 there.
  3. Inverse-tangent values must come from the principal range.

Every fact used, and where it comes from

FactUsed inSource
∫abf=F(b)−F(a)\int_a^b f = F(b) - F(a) for continuous ffSteps 1, 5Theorem 2, Section 7.8.3 (Lesson)
Substitution t=x2+4t = x^2 + 4⁠; ∫dtt=log⁡∣t∣\int\frac{dt}{t} = \log|t|Step 3Section 7.3.1; standard formula (xii), Section 7.2 (Lesson)
∫dxx2+a2=1atan⁡−1xa+C\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a} + CStep 3Formula (3), Section 7.4 (Lesson)
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1 + x^2}Step 4Standard formula (x), Section 7.2
tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, tan⁡−10=0\tan^{-1}0 = 0Step 5Principal values, Chapter 2 (Basics)
log⁡a−log⁡b=log⁡ab\log a - \log b = \log\frac{a}{b}Step 5Laws of logarithms (Basics)
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 →