Integral of x^3sqrt(1+9x^4) dx

1. Let $u = 9 x^{4} + 1$.

   Then let $du = 36 x^{3} dx$ and substitute $\frac{du}{36}$:

   $$\int \frac{\sqrt{u}}{36}\, du$$

   1. The integral of a constant times a function is the constant times the integral of the function:

      $$\int \sqrt{u}\, du = \frac{\int \sqrt{u}\, du}{36}$$

      1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$:

         $$\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}$$

      So, the result is: $\frac{u^{\frac{3}{2}}}{54}$

   Now substitute $u$ back in:

   $$\frac{\left(9 x^{4} + 1\right)^{\frac{3}{2}}}{54}$$

2. Add the constant of integration:

   $$\frac{\left(9 x^{4} + 1\right)^{\frac{3}{2}}}{54}+ \mathrm{constant}$$

The answer is:

$$\frac{\left(9 x^{4} + 1\right)^{\frac{3}{2}}}{54}+ \mathrm{constant}$$