Miscellaneous · Q8: $\int\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}\,dx$

Basics you need

Solution

1. Turn each exponential into a power of $x$

   For $x > 0$ and each $k$⁠: $e^{k\log x} = e^{\log x^k} = x^k$⁠. So

   $$f(x) = \frac{x^5 - x^4}{x^3 - x^2}.$$
   No method of integration applies well to the given form, but the form hides simple powers. Simplifying first is always worth trying before reaching for substitution or parts. The step needs $x > 0$⁠, which is required anyway for $\log x$ to exist.
   $k\log x = \log x^k$ and $e^{\log y} = y$ (Basic 1). Basics: logarithms

2. Factorise, cancel, and record the domain

   $$f(x) = \frac{x^4(x - 1)}{x^2(x - 1)} = x^2$$

   for $x > 0$⁠, $x \ne 1$⁠.

   The denominator $x^2(x - 1)$ is 0 at $x = 1$ (and at $x = 0$⁠, already excluded), so the original function is not defined there. For every other $x > 0$ the common factors $x^2$ and $x - 1$ are non-zero and may be cancelled.
   Cancelling a common factor is allowed only where it is not zero.

3. Integrate with the power rule

   $$\int f(x)\,dx = \int x^2\,dx = \frac{x^3}{3} + C.$$
   Standard formula (i) with $n = 2$⁠. The formula holds on each of the intervals $0 < x < 1$ and $x > 1$⁠, and the constant may differ between them. This is the textbook's answer.
   Standard formula (i), Section 7.2. Basics: the power rule

4. Check: differentiate the answer

   $\dfrac{d}{dx}\Big(\dfrac{x^3}{3}\Big) = x^2$⁠, which equals $f(x)$ at every point of its domain by Step 2. ✓

   Directly: at $x = 2$⁠, $e^{5\log 2} = 2^5 = 32$⁠, and so on, giving $\frac{32 - 16}{8 - 4} = 4 = 2^2$⁠.
   Basics: integration undoes differentiation

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