Answer :
I don't see a square root sign anywhere, so I'll assume the integral is
[tex]\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx[/tex]
First complete the square:
[tex]13+12x-x^2=49-(6-x)^2=7^2-(6-x)^2[/tex]
Now in the integral, substitute
[tex]6-x=7\sin t\implies\mathrm dx=-7\cos t\,\mathrm dt[/tex]
so that
[tex]t=\sin^{-1}\left(\dfrac{6-x}7\right)[/tex]
Under this change of variables, we have
[tex]7^2-(6-x)^2=7^2-7^2\sin^2t=7^2(1-\sin^2t)=7^2\cos^2t[/tex]
so that
[tex]\displaystyle\int\sqrt{13+12x-x^2}\,\mathrm dx=-7\int\sqrt{7^2\cos^2t}\,\cos t\,\mathrm dt=-49\int|\cos t|\cos t\,\mathrm dt[/tex]
Under the right conditions, namely that cos(t) > 0, we can further reduce the integrand to
[tex]|\cos t|\cos t=\cos^2t=\dfrac{1+\cos(2t)}2[/tex]
[tex]\displaystyle-49\int|\cos t|\cos t\,\mathrm dt=-\frac{49}2\int(1+\cos(2t))\,\mathrm dt=-\frac{49}2\left(t+\frac12\sin(2t)\right)+C[/tex]
Expand the sine term as
[tex]\dfrac12\sin(2t)}=\sin t\cos t[/tex]
Then
[tex]t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \sin t=\dfrac{6-x}7[/tex]
[tex]t=\sin^{-1}\left(\dfrac{6-x}7\right)\implies \cos t=\sqrt{7^2-(6-x)^2}=\sqrt{13+12x-x^2}[/tex]
So the integral is
[tex]\displaystyle-\frac{49}2\left(\sin^{-1}\left(\dfrac{6-x}7\right)+\dfrac{6-x}7\sqrt{13+12x-x^2}\right)+C[/tex]