How to calculate $\int{\frac{dx}{3x^2+2}}$? 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.
 A: 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 $3x^2+2 = 2u^2+2 = 2(u^2+1)$?
A: If we put $\sqrt{\frac{3}{2}}x=t$ then it means that $\mathrm dx=\sqrt{\frac{2}{3}}\mathrm dt$
$$\begin{eqnarray*}
\int{\dfrac{\mathrm dx}{3x^2+2}} &=& \displaystyle \frac{1}{2}\int{\dfrac{\mathrm dx}{(\sqrt{\frac{3}{2}}x)^2+1}}\\
&=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}}\int{\dfrac{\mathrm dt}{t^2+1}}\\
&=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}} \arctan(t)+\text C\\
&=&\frac{\sqrt{2}}{2\sqrt{3}}\arctan\left(\sqrt{\frac{3}{2}}x\right)+\text C\\
&=&\frac{1}{\sqrt{6}}\arctan{\left(\sqrt{\frac{3}{2}}x\right)}+\text C
\end{eqnarray*}$$
A: $$\displaystyle\int{\dfrac{dx}{3x^2+2}}$$
put $x =\sqrt{\frac{2}{3}}.\tan \theta $
$$\displaystyle\int{\dfrac{\sqrt{\frac{2}{3}}. (\sec\theta)^2. d\theta}{2+2(\tan\theta)^2}}$$
$$\sqrt\frac{1}{6}\displaystyle\int{d\theta}=\sqrt\frac{1}{6}\theta+c=\sqrt\frac{1}{6}.\arctan{\left(\sqrt{\frac{3}{2}}x\right)}+c$$
