How to calculate

$$\int \frac{x^{\:}}{\sqrt{x^4+3}}\ dx$$

  • 2
    $\begingroup$ Hint: What about the sustitution $x^2=\sqrt{3}\tan t$? $\endgroup$ – Ángel Mario Gallegos Jun 17 '16 at 3:32
  • 1
    $\begingroup$ The trick here is reducing to $$\int \frac{1}{\sqrt{s^2+1}}\,ds$$ $\endgroup$ – user228113 Jun 17 '16 at 3:32
  • $\begingroup$ (assuming the "surface-integral" is just a bad choice of tag) $\endgroup$ – user228113 Jun 17 '16 at 3:34
  • $\begingroup$ @SophieAgnesi The first edit after the question was put on hold gets it into reopen review queue, see meta (if it is within 5 days). For this reason I think that if a question is on hold, we should avoid minor edits. An edit which addresses all reasons why the post was put on hold is preferred. $\endgroup$ – Martin Sleziak Jun 17 '16 at 4:50

Try to show what you think will work or what you've tried next time you ask a question. $$\int \frac{x^{\:}}{\sqrt{x^4+3}}dx$$ $$u=x^2\Rightarrow du=2x~dx$$ $$=\frac{1}{2}\int \dfrac{1}{\sqrt{u^2+3}}du$$ Substituting the integral now $$u=\sqrt{3}\tan v\Rightarrow du=\sqrt{3}\sec^2dv$$ $$=\frac{1}{2}\sqrt{3}\int\dfrac{\sec^2v}{\sqrt{3\tan^2v+3}}dv$$ $$=\frac{1}{2}\sqrt{3}\int\dfrac{\sec^2v}{\sec^2v\sqrt{3}}dv$$ $$=\frac{1}{2}\sqrt{3}\dfrac{1}{\sqrt{3}}\int \sec(v)~dv$$ $$=\frac{1}{2}\sqrt{3}\dfrac{1}{\sqrt{3}}\ln(\tan v+\sec v)$$ Now we substitute $v$ back in $$=\frac{1}{2}\sqrt{3}\dfrac{1}{\sqrt{3}}\ln \left(\tan \left(\arctan\left(\dfrac{1}{\sqrt{3}}x^2\right)\right)+\sec \left(\arctan\left(\dfrac{1}{\sqrt{3}}x^2\right)\right)\right)$$ $$=\frac{1}{2}\ln\left(\sqrt{\dfrac{x^4}{3}+1}+\dfrac{x^2}{\sqrt{3}}\right)+C$$ $$=\dfrac{\ln\left(\dfrac{x^2}{\sqrt{3}}+\sqrt{\dfrac{x^4+3}{3}}\right)}{2}+C$$ $$=\dfrac{\ln\left(\dfrac{x^2+\sqrt{x^4+3}}{\sqrt{3}}\right)}{2}+C$$ $$=\dfrac{\ln\left(x^2+\sqrt{x^4+3}\right)-\dfrac{\ln3}{2}}{2}+C$$ $$=\dfrac{2\ln\left(x^2+\sqrt{x^4+3}\right)-\ln3}{4}+C$$


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