Miscellaneous · Q38 (MCQ): $\int\frac{dx}{e^x+e^{-x}}$

Basics you need

Solution

1. Where is the integrand defined?

   $e^x > 0$ and $e^{-x} > 0$ for every real $x$⁠, so $e^x + e^{-x} > 0$ and the integrand is defined and continuous on all of $\R$⁠.

   The answer we find should therefore also work for every real $x$⁠. That alone casts doubt on option (C), which needs $e^x - e^{-x} > 0$⁠, that is $x > 0$⁠.

2. Multiply top and bottom by $e^x$

   $$\frac{1}{e^x + e^{-x}} = \frac{e^x}{e^x\cdot e^x + e^x\cdot e^{-x}} = \frac{e^x}{e^{2x} + 1} .$$
   The negative power $e^{-x}$ is awkward. Multiplying by $e^x$ (which is never 0) clears it, and as a bonus puts $e^x$ in the numerator, the derivative of the $e^x$ inside $e^{2x} = (e^x)^2$⁠. That is the pattern substitution needs.
   $e^x e^{-x} = 1$⁠, $e^x e^x = e^{2x}$⁠.

3. Substitute $t = e^x$ and integrate

   Put $t = e^x$⁠, $dt = e^x\,dx$⁠. Then

   $$\begin{aligned} \int\frac{e^x\,dx}{(e^x)^2 + 1} &= \int\frac{dt}{t^2 + 1} \\ &= \tan^{-1}t + C \\ &= \tan^{-1}(e^x) + C . \end{aligned}$$
   Standard formula (x), Section 7.2. The substitution is valid for every real $x$⁠: $e^x$ is differentiable with derivative $e^x$⁠, and $t = e^x$ takes only positive values, where $\frac{1}{1+t^2}$ is continuous.
   Lesson: integration by substitution

4. Check: differentiate the answer back

   $$\begin{aligned} \frac{d}{dx}\tan^{-1}(e^x) &= \frac{e^x}{1 + e^{2x}} \\ &= \frac{e^x}{e^x(e^{-x} + e^x)} \\ &= \frac{1}{e^x + e^{-x}} \ ✓ \end{aligned}$$

   valid for every real $x$⁠.

   Chain rule: outer $\tan^{-1}$⁠, with derivative $\frac{1}{1+u^2}$⁠; inner $u = e^x$⁠, with derivative $e^x$⁠. Dividing top and bottom by $e^x$ reverses Step 2.
   Basics: integration undoes differentiation

5. Examine every option

   At $x = 0$ the integrand is $\frac{1}{1+1} = \frac12$⁠; at $x = \log 2$ (where $e^x = 2$⁠, $e^{-x} = \frac12$⁠) it is $\frac{1}{2.5} = 0.4$⁠.

   • (A) Derivative $\frac{e^x}{1+e^{2x}}$⁠, equal to the integrand everywhere (Step 4). At $x = 0$⁠: $\frac12$ ✓; at $x = \log 2$⁠: $\frac{2}{5} = 0.4$ ✓. Correct.
   • (B) $\frac{d}{dx}\tan^{-1}(e^{-x})$ $= \frac{-e^{-x}}{1 + e^{-2x}}$ $= -\frac{1}{e^x + e^{-x}}$ (multiply top and bottom by $e^x$⁠). This is the negative of the integrand: $-\frac12$ at $x = 0$⁠. Wrong. (In fact (B) is minus (A) plus a constant: the sum of $\tan^{-1}(e^x)$ and $\tan^{-1}(e^{-x})$ has derivative 0 on all of $\R$⁠, so it is constant, and at $x = 0$ it is $2\tan^{-1}1 = \frac{\pi}{2}$⁠. So $\tan^{-1}(e^{-x})$ $= \frac{\pi}{2} - \tan^{-1}(e^x)$⁠.)
   • (C) $\log(e^x - e^{-x})$ is defined only for $x > 0$⁠, while the integrand is defined everywhere. Its derivative is $\frac{e^x + e^{-x}}{e^x - e^{-x}}$⁠; at $x = \log 2$ this is $\frac{2.5}{1.5} = \frac53 \ne 0.4$⁠. Wrong.
   • (D) Derivative $\frac{e^x - e^{-x}}{e^x + e^{-x}}$⁠: at $x = 0$ it is $0 \ne \frac12$⁠; at $x = \log 2$ it is $\frac{1.5}{2.5} = 0.6 \ne 0.4$⁠. Wrong. (It would be right for the integrand $\frac{e^x - e^{-x}}{e^x + e^{-x}}$⁠, a different question.)
   (C) and (D) come from treating the denominator as if its derivative were in the numerator. It is not: the numerator here is 1.

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