Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I've started doing $$\displaystyle\int{\dfrac{dx}{3x^2+2}}$$ but I only get $$\displaystyle\int{(3x^2+2)^{-1}dx}\\ \frac{1}{6}\displaystyle\int{\frac{6x(3x^2+2)^{-1}}{x}dx}\\ $$ And I don't know how to do solve this.

share|cite|improve this question
Do you know residue theorem? – Argon Apr 27 '12 at 20:03
@Argon Why would you need residues here? – Pedro Tamaroff Apr 27 '12 at 20:22
@PeterTamaroff I find them easier to compute this integral then trigonometric substitutions. – Argon Apr 27 '12 at 20:27
@Argon But the OP is asking for a primitive, not a definite integral. How do residues play a role here? – Pedro Tamaroff Apr 27 '12 at 20:43
@Garmen: There was a conflation of definite and indefinite integrals; currently, there is an error in not changing the interval of integration when doing a change of variable (in the second step). If you change everything to indefinite integrals and drop the interval $[a,b]$, then the result is incomplete/incorrect for two reason: (i) the problem is about a function of $x$, the answer is about a function of $t$; and (ii) it would be missing the constant of integration. – Arturo Magidin Apr 27 '12 at 20:52
up vote 2 down vote accepted

If we put $\sqrt{\frac{3}{2}}x=t$ then it means that $\mathrm dx=\sqrt{\frac{2}{3}}\mathrm dt$

$$\begin{eqnarray*} \int{\dfrac{\mathrm dx}{3x^2+2}} &=& \displaystyle \frac{1}{2}\int{\dfrac{\mathrm dx}{(\sqrt{\frac{3}{2}}x)^2+1}}\\ &=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}}\int{\dfrac{\mathrm dt}{t^2+1}}\\ &=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}} \arctan(t)+\text C\\ &=&\frac{\sqrt{2}}{2\sqrt{3}}\arctan\left(\sqrt{\frac{3}{2}}x\right)+\text C\\ &=&\frac{1}{\sqrt{6}}\arctan{\left(\sqrt{\frac{3}{2}}x\right)}+\text C \end{eqnarray*}$$

share|cite|improve this answer
@Arturo and Abdelmajid: due to the large number of substantial edits, the answer has been converted to Community Wiki. I'm also purging all comments since it appears that they are no longer relevant. – Willie Wong Apr 30 '12 at 8:45

Hint. Do you know how to integrate $$\int\frac{1}{u^2+1}\,du\ ?$$ If so, can you make a change of variable, say $u=kx$ for some constant $k$, so that $3x^2+2 = 2u^2+2 = 2(u^2+1)$?

share|cite|improve this answer
For integration like this, if you can imagine a "sum of squares" or "difference of squares" it could suggest a right triangle, and then a trigonometric substitution. – GEdgar Apr 27 '12 at 19:54


put $x =\sqrt{\frac{2}{3}}.\tan \theta $

$$\displaystyle\int{\dfrac{\sqrt{\frac{2}{3}}. (\sec\theta)^2. d\theta}{2+2(\tan\theta)^2}}$$


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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