Both integrals can be done by parts.
For the first one:
[tex]\displaystyle \int (x^3-7x)\ln(x) \,\mathrm dx = uv - \int v\,\mathrm du[/tex]
where
[tex]u = \ln(x) \implies \mathrm du = \dfrac{\mathrm dx}x \\\\ \mathrm dv = (x^3-7x)\,\mathrm dx \implies v = \dfrac{x^4}4-\dfrac{7x^2}2[/tex]
Then
[tex]\displaystyle \int (x^3-7x)\ln(x)\,\mathrm dx = \left(\dfrac{x^4}4-\dfrac{7x^2}2\right)\ln(x) - \int\left(\dfrac{x^4}4-\dfrac{7x^2}2\right)\dfrac{\mathrm dx}x \\\\ \int (x^3-7x)\ln(x)\,\mathrm dx = \left(\dfrac{x^4}4-\dfrac{7x^2}2\right)\ln(x) - \int\left(\dfrac{x^3}4-\dfrac{7x}2\right)\,\mathrm dx \\\\ \int (x^3-7x)\ln(x)\,\mathrm dx = \boxed{\left(\dfrac{x^4}4-\dfrac{7x^2}2\right)\ln(x) - \dfrac{x^4}{16} + \dfrac{7x^2}4 + C}[/tex]
You can treat the second one identically, and you would end up with
[tex]\displaystyle \int(x^2-1)\ln(x)\,\mathrm dx = \left(\frac{x^3}3-x\right)\ln(x) - \int\left(\frac{x^2}3-1\right)\,\mathrm dx \\\\ \int(x^2-1)\ln(x)\,\mathrm dx = \boxed{\left(\frac{x^3}3-x\right)\ln(x) - \frac{x^3}9+\frac{x^2}2} + C}[/tex]
More generally, the antiderivative of a polynomial [tex]p(x)[/tex] multiplied by [tex]\ln(x)[/tex] is given by
[tex]\displaystyle \int p(x)\ln(x)\,\mathrm dx = \ln(x)\int p(x)\,\mathrm dx - \int\frac{\displaystyle \int p(u)\,\mathrm du}x\,\mathrm dx[/tex]
