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In Mathematics, we know the following is true:

$$\int \frac{1}{x} \space dx = \ln(x)$$

Not only that, this rule works for constants added to x: $$\int \frac{1}{x + 1}\space dx = \ln(x + 1) + C{3}$$ $$\int \frac{1}{x + 3}\space dx = \ln(x + 3) + C$$ $$\int \frac{1}{x - 37}\space dx = \ln(x - 37) + C$$ $$\int \frac{1}{x - 42}\space dx = \ln(x - 42) + C$$

So its pretty safe to say that $$\int \frac{1}{x + a}\space dx = \ln(x + a) + C$$ But the moment I introduce $x^a$ where $a$ is not equal to 1, the model collapses. The integral of $1/x^a$ is not equal to $\ln(x^a)$. The same goes for $\cos(x)$, and $\sin(x)$, and other trig functions.

So when are we allowed or not allowed to use the rule of $\ln(x)$ when integrating functions?

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Just a note: $\int \frac{1}{x} \space dx = \ln|x|$ –  Joe Jun 19 '12 at 22:20
Oh, of course! I always forget that. Will edit. –  Zolani13 Jun 19 '12 at 22:20
Whenever it is true. It's a mnemonic, not an actual rule or definition. –  tomasz Jun 19 '12 at 22:20
Also, it doesn't work for $ax+b$. $\int 1/(ax+b)=\log\lvert x+b/a\rvert /a$. –  tomasz Jun 19 '12 at 22:23

3 Answers 3

up vote 2 down vote accepted

Generally speaking, "using $\ln (x)$" as a rule or technique is unheard of. When one speaks of techniques, they usually include integration by substitution, integration by parts, trig substitutions, partial fractions, etc. With introductory calculus in mind, $\ln |x|$ is defined as $\int \frac{1}{x} \ dx.$ This can be extended to $\ln |u| = \int \frac{1}{u} \ du.$ Note that there are many more definitions for $\ln (x)$, but I felt this best related particularly to your examples.

For your first couple of examples, when choosing your $u$ to be the denominator, the $du$ is simply equal to $dx.$ This is what 'allows' the integrand to be evaluated to just $\ln |u|$ where $u$ is a linear expression.

In regards to $\int \frac{1}{x^2 + a} \ dx$, this can be handled using an inverse tangent and would be evaluated to $$\dfrac{\operatorname{arctan}(\frac{x}{\sqrt{a}})}{\sqrt{a}} + C$$

For integrals of the form $$\int \frac{1}{x^n + a} \ dx$$

where $n \ge 3$, you will have to revert to partial fractions. For more on partial fractions, see this.

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It boils down to $u$-substitution (which if you haven't covered yet, you soon will). You know that $$ \int \frac{1}{x} dx = \ln x + C $$ (I'll not bother with absolute values here. You should remember them on problems that you do for class, but they aren't the focus of this question.)

Now, to handle $$ \int \frac{1}{ax+b}dx $$ we do a $u$-substitution with $u = ax+b$. This makes $du=adx$, so the integral becomes $$ \int \frac{1}{u} \frac{du}{a} = \frac{1}{a} \ln u + C = \frac{1}{a}\ln(ax+b) + C$$

Note the result: $$ \int \frac{1}{ax+b}dx = \frac{1}{a}\ln(ax+b) + C $$

Check it by taking a derivative of the right-hand side. A chain rule was needed to take that derivative. The $u$-substitution we did was the chain rule reversed.

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Perhaps I can reverse-address your question. Oftentimes (typically in optimization problems) when dealing with a positive real function $f$ it is easier to differentiate $\log f$ than $f$ itself. It's easy to check that the so-called logarithmic derivative satisfies $\frac{d}{dx} \log [f (x)] = \frac{f'(x)}{f(x)}.$ Note also that we can recover the original derivative by multiplying through with $f.$ In terms of primitives, this is the same as saying $\displaystyle\int \frac{f'(x)}{f(x)} dx = \log x + C.$ This is a general case of some of the formulae you presented and is useful in quickly evaluating many definite integrals by substitution - for instance, $\displaystyle\int \cot x $ $dx = \log |\sin x | +C$ is immediate by this formula.

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