Calculating 2 non-definite integrals Calculate the following:
A) $\int \sqrt{3x^4 +x^6 +9x^2} \, dx$
B) $\int \sqrt[3]{{\frac{1}{x^2 +1}}}\, dx$
A) I managed to write $\int x \sqrt{3x^2 +x^4 +9} \, dx$, but then I didn't know what to do because of the square root, even with integration by parts.
B) I tried substituting $u=x^2 +1$ so $du=2x\,dx$ but then I don't have any $x$ for $du$.

In the integral
$$\int x \sqrt{x^4+3x^2+9} \ dx$$
substitute $u=x^2$, $du=2xdx$ to get
$$\frac{1}{2} \int \sqrt{u^2+3u+9} \ du = \frac{1}{2} \int \sqrt{ \left( u+\frac{3}{2} \right)^2 + \frac{27}{4} } \ du$$ and to continue, substitute
$$u+\frac{3}{2}= \frac{3\sqrt 3}{2} \tan \theta$$
you will need to remember the integral of $\sec^3 \theta$.
