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Chapter 7 · Integrals

Exercise 7.2 · Q18: ∫etan⁡−1x1+x2 dx\int\frac{e^{\tan^{-1}x}}{1+x^2}\,dx

Exercise 7.2Integration by substitution · Section 7.3.1
The question

Integrate the function

etan⁡−1x1+x2.\frac{e^{\tan^{-1}x}}{1 + x^2}.
Exam answer

Let I=∫etan⁡−1x1+x2 dxI = \displaystyle\int\frac{e^{\tan^{-1}x}}{1 + x^2}\,dx⁠.

Put tan⁡−1x=t\tan^{-1}x = t⁠. Then dx1+x2=dt\frac{dx}{1 + x^2} = dt⁠.

I=∫et dt=et+CI = \int e^t\,dt = e^t + C

∴ I=etan⁡−1x+C\therefore\ I = e^{\tan^{-1}x} + C⁠.

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

The one idea this problem is built on

The exponent is tan⁡−1x\tan^{-1}x⁠, and the factor 11+x2\frac{1}{1+x^2} is exactly its derivative. So put t=tan⁡−1xt = \tan^{-1}x⁠: then dt=dx1+x2dt = \frac{dx}{1+x^2}⁠, and the whole integral becomes ∫et dt=et+C\int e^t\,dt = e^t + C⁠.

Given, to find, and the plan
Given
f(x)=etan⁡−1x1+x2f(x) = \dfrac{e^{\tan^{-1}x}}{1+x^2}⁠. Here tan⁡−1x\tan^{-1}x is defined for every real xx and 1+x2≥1>01 + x^2 \ge 1 > 0⁠, so ff is defined everywhere.
To find
∫etan⁡−1x1+x2 dx\displaystyle\int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx⁠.
Plan
(1) Recognise 11+x2\frac{1}{1+x^2} as the derivative of tan⁡−1x\tan^{-1}x⁠. (2) Substitute t=tan⁡−1xt = \tan^{-1}x⁠. (3) Integrate ete^t⁠. (4) Return to xx⁠. (5) Differentiate back.

Basics you need

1The derivative of tan⁡−1x\tan^{-1}x

ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2} for every real xx (Section 5.3.3; also formula (x) of Section 7.2). The principal value tan⁡−1x\tan^{-1}x lies in (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠. Basics: integration undoes differentiation

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

3The exponential integral

∫et dt=et+C\int e^t\,dt = e^t + C (standard formula (xi), Section 7.2).

Solution

  1. Spot the inner function and its derivative

    Write the integrand as

    etan⁡−1x⋅11+x2.e^{\tan^{-1}x}\cdot\frac{1}{1+x^2}.

    The inner function is g(x)=tan⁡−1xg(x) = \tan^{-1}x⁠, and g′(x)=11+x2g'(x) = \frac{1}{1+x^2} is the second factor.

    This is the pattern f(g(x)) g′(x)f(g(x))\,g'(x) with outer function f(t)=etf(t) = e^t⁠. Because the derivative appears exactly (not even a constant factor is needed), the substitution will be clean.
    Section 5.3.3, derivatives of inverse trigonometric functions. Basics: integration undoes differentiation

  2. Substitute t=tan⁡−1xt = \tan^{-1}x

    Let t=tan⁡−1xt = \tan^{-1}x⁠. Then

    dt=dx1+x2.dt = \frac{dx}{1+x^2}.

    As xx runs over all real numbers, tt runs over the interval (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)⁠. The integral becomes

    ∫etan⁡−1x dx1+x2=∫et dt.\int e^{\tan^{-1}x}\,\frac{dx}{1+x^2} = \int e^t\,dt.
    tan⁡−1\tan^{-1} is differentiable at every real number, so the substitution is valid on the whole real line. Every xx has disappeared from the integral, so the substitution is complete. This is the same substitution the book uses in Example 5(iv).
    Change of variable, Section 7.3.1. Lesson: integration by substitution

  3. Integrate and return to xx

    ∫et dt=et+C=etan⁡−1x+C.\int e^t\,dt = e^t + C = e^{\tan^{-1}x} + C.
    Standard formula (xi): ete^t is its own antiderivative. Replacing tt by tan⁡−1x\tan^{-1}x gives the answer in terms of xx⁠.
    Standard formula (xi), Section 7.2.

  4. Check: differentiate the answer back

    By the chain rule (outer eue^u⁠, inner u=tan⁡−1xu = \tan^{-1}x⁠):

    ddxetan⁡−1x=etan⁡−1x⋅11+x2=etan⁡−1x1+x2.\frac{d}{dx}e^{\tan^{-1}x} = e^{\tan^{-1}x}\cdot\frac{1}{1+x^2} = \frac{e^{\tan^{-1}x}}{1+x^2}.

    This is the integrand, so the answer is correct for every real xx⁠.

    The chain rule multiplies by the derivative of the inner function, 11+x2\frac{1}{1+x^2}⁠, and that reproduces the second factor of the integrand. Getting the integrand back confirms the answer (Property (I), Section 7.2.1).
    Chain rule, Theorem 4, Section 5.3.1; ddxex=ex\frac{d}{dx}e^x = e^x⁠, Theorem 5, Section 5.4.
Result ∫etan⁡−1x1+x2 dx=etan⁡−1x+C,x∈R.\int \frac{e^{\tan^{-1}x}}{1+x^2}\,dx = e^{\tan^{-1}x} + C, \qquad x \in \R.

(Textbook answer: etan⁡−1x+Ce^{\tan^{-1}x} + C⁠.)

Check it with numbers

Let F(x)=etan⁡−1xF(x) = e^{\tan^{-1}x} and look near x=1x = 1 with h=0.001h = 0.001⁠:

  • F(1.001)≈2.194376417F(1.001) \approx 2.194376417 and F(0.999)≈2.192183137F(0.999) \approx 2.192183137⁠.
  • Slope estimate: F(1.001)−F(0.999)0.002\dfrac{F(1.001) - F(0.999)}{0.002} ≈1.096640\approx 1.096640⁠.
  • Integrand at x=1x = 1⁠: tan⁡−11=π4\tan^{-1}1 = \frac{\pi}{4}⁠, so it is eπ/42≈1.096640\dfrac{e^{\pi/4}}{2} \approx 1.096640⁠.

They agree.

Common mistakes
  1. Integrating the two factors separately. The integral of a product is not the product of the integrals, so an answer such as etan⁡−1x⋅tan⁡−1x+Ce^{\tan^{-1}x}\cdot\tan^{-1}x + C is wrong (differentiate it and the product rule produces extra terms). The factor 11+x2\frac{1}{1+x^2} is not integrated on its own; it is absorbed into dtdt⁠.
  2. Reading tan⁡−1x\tan^{-1}x as 1tan⁡x\frac{1}{\tan x}⁠. Here it is the inverse tangent, whose derivative is 11+x2\frac{1}{1+x^2}⁠; 1tan⁡x\frac{1}{\tan x} is cot⁡x\cot x⁠.
  3. Leaving tt in the answer. Substitute back t=tan⁡−1xt = \tan^{-1}x⁠.
What to take away
  1. 11+x2\frac{1}{1+x^2} next to a function of tan⁡−1x\tan^{-1}x is a signal for t=tan⁡−1xt = \tan^{-1}x⁠. Likewise 11−x2\frac{1}{\sqrt{1-x^2}} next to a function of sin⁡−1x\sin^{-1}x signals t=sin⁡−1xt = \sin^{-1}x (Question 23).
  2. ∫eg(x) g′(x) dx=eg(x)+C\displaystyle\int e^{g(x)}\,g'(x)\,dx = e^{g(x)} + C⁠.

Every fact used, and where it comes from

FactUsed inSource
ddxtan⁡−1x=11+x2\frac{d}{dx}\tan^{-1}x = \frac{1}{1+x^2}Steps 1, 2, 4Section 5.3.3; formula (x), Section 7.2
Principal range of tan⁡−1\tan^{-1} is (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)Step 2Chapter 2, Section 2.2
Change of variable (substitution)Step 2Section 7.3.1; Example 5(iv) (Lesson: integration by substitution)
∫et dt=et+C\int e^t\,dt = e^t + CStep 3Standard formula (xi), Section 7.2
ddxex=ex\frac{d}{dx}e^x = e^x⁠; chain ruleStep 4Theorem 5, Section 5.4; Theorem 4, Section 5.3.1
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