## Calculus, 10th Edition (Anton)

By using $u=ln(x)$ and $dv=x^{3}$, $du=\frac{dx}{x}$ and $v=\frac{1}{4}x^{4}$. This means that $$\int{udv}=uv-\int{vdu}=\frac{1}{4}x^{4}ln(x)-\int{\frac{x^{4}}{4x}}dx$$ The newly formed integral $\int{vdu}$ can now be solved using power rule after dividing $x^4$ by $4x$.