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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.

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

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

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@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
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$$\displaystyle\int{\dfrac{dx}{3x^2+2}}$$

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

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

$$\sqrt\frac{1}{6}\displaystyle\int{d\theta}=\sqrt\frac{1}{6}\theta+c=\sqrt\frac{1}{6}.\arctan{\left(\sqrt{\frac{3}{2}}x\right)}+c$$

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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 $3x^2+2 = 2u^2+2 = 2(u^2+1)$?

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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
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