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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we have: if we put $\sqrt{\frac{3}{2}}x=t$ then we have $dx=\sqrt{\frac{2}{3}}dt$ $$\begin{eqnarray*} \int{\dfrac{dx}{3x^2+2}} &=& \displaystyle \frac{1}{2}\int{\dfrac{dx}{(\sqrt{\frac{3}{2}}x)^2+1}}\\ &=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}}\int{\dfrac{dt}{t^2+1}}\\ &=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}} \arctan(t)+C\\ &=&\frac{\sqrt{2}}{2\sqrt{3}}\arctan\left(\sqrt{\frac{3}{2}}x\right)+C\\ &=&\frac{1}{\sqrt{6}}\arctan{\left(\sqrt{\frac{3}{2}}x\right)}+C \end{eqnarray*}$$ |
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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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$$\displaystyle\int{\dfrac{dx}{3x^2+2}}$$ 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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