Integrate $\int \frac {1+x^2}{\sqrt{6+3x^2}}dx$

Question

I can't find the integral $$\int \frac {1+x^2}{\sqrt{6+3x^2}}dx$$ I can't figure out the u-substitution to use, but I also can't really deal with trigonometric substitutions. I tried to substitution twice and DI method. I may have made a mistake, but couldn't figure it out. In DI method, I differentiated the numerator and integrated the denominator.

I found that the problem lied on the integration of the denominator as I can't really factor it so I couldn't use partial fractions. Please help.

Answer 1

Substitute $y= \sqrt{6+3x^2}$. Then, $ydy =3x dx$ and \begin{align} \int \frac {1+x^2}{\sqrt{6+3x^2}}dx &= \int \frac {1+x^2}{y}dx= \frac16\int \frac {y^2 +3x^2}{y}dx\\ &=\frac16\int (ydx +xdy)=\frac16\int d(xy) =\frac16xy+C \end{align}

Answer 2

Let $x=\sqrt2\tan\theta$, $dx=\sqrt2\sec^2\theta$, the integral becomes $$\frac{1}{\sqrt3}\int(2\sec^3\theta-\sec\theta)d\theta=\frac{1}{\sqrt3}\sec\theta\tan\theta+c=\frac{1}{2\sqrt3}x\sqrt{x^2+2}+c$$ since $\tan\theta=\frac{x}{\sqrt2}$ and $\sec\theta=\frac{\sqrt{x^2+2}}{\sqrt2}.$