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 $du$.
On
The two integrals don't appear to be related, so it would've been better to ask them as separate questions. At any rate:
(A) Substituting $u = x^2, du = 2 x \,dx$ transforms the integral to $$\frac12 \int \sqrt{u^2 + 3 u + 9} \,du .$$ Applying the hyperbolic substitution $$u + \frac{3}{2} = \frac{3 \sqrt 3}{2} \sinh t, \qquad du = \frac{3 \sqrt 3}{2} \cosh t \,dt$$ transforms the integral to $$\frac{27}{8} \int \cosh^2 t \,dt = \frac{27}{8} \int \frac12 (1 + \cosh 2 t) \,dt = \frac{27}{16} \left(t + \frac12 \sinh 2 t\right) + C .$$
(B) There doesn't seem to be a closed-form value for this indefinite integral in terms of elementary functions. In terms of the hypergeometric function ${}_2 F_1$, we have $$\int \frac{dx}{\sqrt[3]{1 + x^2}} = x {}_2 F_1 \left(\frac12, \frac13; \frac32; -x^2\right) + C .$$ More generally, $$\int (1 + x^2)^a \,dx = x {}_2 F_1 \left(\frac12, -a; \frac32; -x^2\right) + C .$$
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 $\sec^3 \theta$.