When integrating $A/x$, why use the logarithm instead of $x$ raised to a power?

$\int\frac{15}{x}dx$ would be $$15\int\frac{1}{x}dx = 5\ln|x|+c$$

This seems like a silly question but I'm feeling exceptionally dense today. Why would you apply the logarithm rule, why wouldn't raising $x$ to a $-1$ exponent work?

• Your solution is correct. $\int x^tdx=\frac{1}{t+1}x^{t+1} +C$ is valid for $t\neq -1$, so doesn't apply here. – vadim123 Jun 2 '13 at 4:06
• @vadim123 Ok, i see, thanks! – Ellen Jun 2 '13 at 4:10
• As to "why," if you increment the power of $x^{-1}$ you should find something isn't right since you'll be left with $x^0=1$, whose derivative is not $x^{-1}$ (and which will actually give you 0 area for any bounds). – Zen Jun 2 '13 at 4:21
• Also if you try to take the antiderivative of $x^{-1}$ using the power rule, you would obtain $\dfrac{x^0}{0}$. This is undefined, since the expression divides by zero. – Adriano Jun 2 '13 at 4:44

Let $a$ be a fixed positive number. We know by integration that $$\int_1^a \frac{dx}{x}=\ln a-\ln 1=\ln a.\tag{1}$$
We try to reconcile this with the ordinary formula for the integral of a power. Note that if $w\ne 1$, then $$\int_1^a x^{-w}\,dx= \frac{1}{1-w}\left(a^{1-w}-1^{1-w}\right).\tag{2}$$ There is an obvious problem in using this when $w=1$, because of the division by $0$. So instead let us take the limit as $w$ approaches $1$ of the expression at the right side of (2). So, letting $h=1-w$, we want to find $$\lim_{h\to 0}\frac{a^h -1}{h}.\tag{3}$$ We recognize the limit (3) as the derivative of the function $a^t$ with respect to $t$, at $t=0$. Simce $a^t=e^{(\ln a)t}$, the derivative at $t=0$ is $\ln a$.
This is the same as the answer (1) that we obtained by using the standard formula for $\int \frac{dx}{x}$. So this formula can be thought of as a limiting case of the 'usual" formula (2).