A bit blurry, but if I'm making it out correctly, that's
[tex]\displaystyle\int t\sqrt{(t^2-9)^3}\,\mathrm dt=\int t(t^2-9)^{3/2}\,\mathrm dt[/tex]
Set [tex]u=t^2-9[/tex]. Then [tex]\mathrm du=2t\,\mathrm dt[/tex], or [tex]t\,\mathrm dt=\dfrac{\mathrm du}2[/tex]. The integral is then equivalent to
[tex]\displaystyle u^{3/2}\dfrac{\mathrm du}2=\frac12\int u^{3/2}\,\mathrm du[/tex]
[tex]=\dfrac12\dfrac{u^{5/2}}{\frac52}+C[/tex]
[tex]=\dfrac12\times\dfrac25u^{5/2}+C[/tex]
[tex]=\dfrac15u^{5/2}+C[/tex]
[tex]=\dfrac15(t^2-9)^{5/2}+C[/tex]
