Looks like the integral is
[tex]I=\displaystyle\int (4x-6)(x^2-3x+1)^5\,\mathrm dx[/tex]
If you we substitute [tex]y=x^2-3x+1[/tex] and [tex]\mathrm dy=(2x-3)\,\mathrm dx[/tex], we get
[tex]I=\displaystyle2\int(2x-3)(x^2-3x+1)^5\,\mathrm dx=2\int y^5\,\mathrm dy[/tex]
Now use the power rule, and replace y accordingly:
[tex]I=\dfrac{2y^6}6+C=\dfrac{y^6}3+C=\boxed{\dfrac{(x^2-3x+1)^5}3+C}[/tex]
