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?

  • 1
    $\begingroup$ Just a note: $\int \frac{1}{x} \space dx = \ln|x|$ $\endgroup$ – Joe Jun 19 '12 at 22:20
  • $\begingroup$ Oh, of course! I always forget that. Will edit. $\endgroup$ – Zolani13 Jun 19 '12 at 22:20
  • 2
    $\begingroup$ Whenever it is true. It's a mnemonic, not an actual rule or definition. $\endgroup$ – tomasz Jun 19 '12 at 22:20
  • 1
    $\begingroup$ Also, it doesn't work for $ax+b$. $\int 1/(ax+b)=\log\lvert x+b/a\rvert /a$. $\endgroup$ – tomasz Jun 19 '12 at 22:23

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.


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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.