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I can't think of a reason why this may be false.

If I have a linear function $f(x)$ which is strictly increasing (assume $f(x) \neq 0$ and NOT OTHER DISCONTINUTIES ) on a domain $D$, then

$\int_D \dfrac{1}{f(x)} dx = \ln(f(x)) \bigg|_{D}$

No absolute value signs because f(x) is monotonically increasing.

I know it's a rather big claim and I tested it out on $\dfrac{1}{1+x}$ and other similarly simple cases and it see to work.

EDIT: I just noticed $\dfrac{1}{1+x}$ also happens to be monotonically increasing on $(0,\infty)$

EDIT2: Let's tweak the conditions on $D$. Assume x > 0

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Consider $f(x) = x/2$. – Karolis Juodelė Jun 2 '12 at 18:53
I am going to add the condition in which for D is x > 0 – Hawk Jun 2 '12 at 18:57
$D$ doesn't matter. My point is that $\int 2/x = 2 \int 1/x$ but $\ln(x/2) \neq 2\ln(x)$ – Karolis Juodelė Jun 2 '12 at 19:02
More generally, $\int{dx\over ax+b}={1\over a}\ln|ax+b|+C$; so your claim is true iff $a=1$. – David Mitra Jun 2 '12 at 19:03
Sorry, what do you mean by "not other discontinuities"? Is $D$ supposed to be a bounded closed interval? – Dylan Moreland Jun 2 '12 at 19:09
up vote 2 down vote accepted

$f(x)=\frac{x}{2}>0$ on $[1,2]$ and is strictly increasing, but $\int_1^2\frac{2}{x}\,dx=\ln(4)\ne \ln(2)=\ln(1)-\ln(1/2).$

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$f$ was supposed to be linear. – Karolis Juodelė Jun 2 '12 at 19:00
Sorry, I Fixed it. – JLA Jun 2 '12 at 19:09

$$\frac{d}{dx}\left( \ln(f(x)) \right) = \frac{f'(x)}{f(x)}.$$

When is this equal to $\dfrac{1}{f(x)}$?

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