Exercise 7.2 · Q19: $\int\frac{e^{2x}-1}{e^{2x}+1}\,dx$

Basics you need

Solution

1. Divide the top and the bottom by $e^x$

   Since $e^x > 0$⁠, dividing numerator and denominator by it does not change the fraction:

   $$\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{e^{x} - e^{-x}}{e^{x} + e^{-x}}.$$
   With $t = e^{2x} + 1$ we would get $dt = 2e^{2x}\,dx$⁠, which does not match the numerator $e^{2x} - 1$⁠. Dividing by $e^x$ makes the expression symmetric, and then the derivative of the denominator, $e^x - e^{-x}$⁠, is exactly the numerator.
   $\frac{e^{2x}}{e^x} = e^{2x - x} = e^x$ and $\frac{1}{e^x} = e^{-x}$⁠. Basics: powers

2. Substitute $t = e^x + e^{-x}$

   Let $t = e^x + e^{-x}$⁠. Then

   $$dt = \left(e^x - e^{-x}\right)dx,$$

   and the integral becomes

   $$\int \frac{e^x - e^{-x}}{e^x + e^{-x}}\,dx = \int\frac{dt}{t}.$$
   The derivative of $e^{-x}$ is $-e^{-x}$ (chain rule with inner function $-x$⁠), which is where the minus sign in $dt$ comes from. That product $\left(e^x - e^{-x}\right)dx$ is precisely the numerator times $dx$⁠.
   Change of variable, Section 7.3.1; Theorem 5, Section 5.4. Lesson: integration by substitution

3. Integrate and return to $x$

   $$\int\frac{dt}{t} = \log|t| + C = \log\left(e^x + e^{-x}\right) + C.$$
   The modulus can be dropped because $t = e^x + e^{-x}$ is a sum of two positive numbers, so $t > 0$ for every $x$⁠. (In fact $t \ge 2$⁠, with equality at $x = 0$⁠.)
   Standard formula (xii), Section 7.2: $\int\frac{dt}{t} = \log|t| + C$⁠. Basics: logarithms and 1/x

4. Check: differentiate the answer back

   For a positive $u$⁠, $\frac{d}{dx}\log u = \frac{u'}{u}$⁠. With $u = e^x + e^{-x}$⁠, $u' = e^x - e^{-x}$⁠:

   $$\begin{aligned} \frac{d}{dx}\log\left(e^x + e^{-x}\right) &= \frac{e^x - e^{-x}}{e^x + e^{-x}} \\ &= \frac{e^{2x} - 1}{e^{2x} + 1}. \end{aligned}$$

   The last line multiplies top and bottom by $e^x$⁠, undoing Step 1. This is the integrand, valid for every real $x$⁠.

   The derivative first comes out in the rewritten form of Step 1; multiplying top and bottom by $e^x > 0$ returns the form printed in the question. Getting the integrand back confirms the answer (Property (I), Section 7.2.1).
   Chain rule, Theorem 4, Section 5.3.1; $\frac{d}{dx}\log x = \frac1x$⁠, Theorem 5, Section 5.4.

5. A second route gives the same answer

   Write the numerator as $2e^{2x} - (e^{2x} + 1)$⁠:

   $$\frac{e^{2x} - 1}{e^{2x} + 1} = \frac{2e^{2x}}{e^{2x} + 1} - 1.$$

   In the first fraction the numerator $2e^{2x}$ is the derivative of $e^{2x} + 1$⁠, so

   $$\int\frac{e^{2x} - 1}{e^{2x} + 1}\,dx = \log\left(e^{2x} + 1\right) - x + C.$$

   This is the same function as before, since $x = \log e^x$⁠:

   $$\begin{aligned} \log\left(e^{2x} + 1\right) - x &= \log\frac{e^{2x} + 1}{e^x} \\ &= \log\left(e^x + e^{-x}\right). \end{aligned}$$
   Two different methods landing on identical functions is a strong check. Here the two forms are equal exactly, not just up to a constant.
   $\log a - \log b = \log\frac{a}{b}$ for $a, b > 0$⁠; $\log e^x = x$⁠. Basics: logarithms

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