It depends on what you mean by the delimiting carats "^"...
Since you use parentheses appropriately in the answer choices, I'm going to go out on a limb here and assume something like "^x^" stands for [tex]\sqrt x[/tex].
In that case, you want to find the antiderivative,
[tex]\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}[/tex]
Complete the square in the denominator:
[tex]9-8x-x^2=25-(16+8x+x^2)=5^2-(x+4)^2[/tex]
Now substitute [tex]x+4=5\sin y[/tex], so that [tex]\mathrm dx=5\cos y\,\mathrm dy[/tex]. Then
[tex]\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\int\frac{5\cos y}{\sqrt{5^2-(5\sin y)^2}}\,\mathrm dy[/tex]
which simplifies to
[tex]\displaystyle\int\frac{5\cos
y}{5\sqrt{1-\sin^2y}}\,\mathrm dy=\int\frac{\cos y}{\sqrt{\cos^2y}}\,\mathrm dy[/tex]
Now, recall that [tex]\sqrt{x^2}=|x|[/tex]. But we want the substitution we made to be reversible, so that
[tex]x+4=5\sin y\iff y=\sin^{-1}\left(\dfrac{x+4}5\right)[/tex]
which implies that [tex]-\dfrac\pi2\le y\le\dfrac\pi2[/tex]. (This is the range of the inverse sine function.)
Under these conditions, we have [tex]\cos y\ge0[/tex], which lets us reduce [tex]\sqrt{\cos^2y}=|\cos y|=\cos y[/tex]. Finally,
[tex]\displaystyle\int\frac{\cos y}{\cos y}\,\mathrm dy=\int\mathrm dy=y+C[/tex]
and back-substituting to get this in terms of [tex]x[/tex] yields
[tex]\displaystyle\int\frac{\mathrm dx}{\sqrt{9-8x-x^2}}=\sin^{-1}\left(\frac{x+4}5\right)+C[/tex]
