Answer
$4-\sqrt 7\approx1.35425$
Work Step by Step
The problem can be represented by equation $\int\limits_4^5\frac{x}{\sqrt {x^2-9}}*dx$
which we can use u-substitution and the substitution rule for definite integrals on.
$u=x^2-9$
$dx=\frac{du}{2x}$
lower limit 4 to $(4)^2-9=7$
upper limit 5 to $(5)^2-9=16$
Now rewrite the equation with these values...
$\int\limits_7^{16}\frac{x}{\sqrt u}*\frac{du}{2x}=\int\limits_7^{16}\frac{1}{2\sqrt u}*du=\int\limits_7^{16}\frac{1}{2}u^{-\frac{1}{2}}*du$
Before our next step we'll have to find the antiderivative of $\frac{1}{2}u^{-\frac{1}{2}}$, which is $u^{\frac{1}{2}}+c$
We can now use the fundamental theorem of calculus.
$u^{\frac{1}{2}}|_7^16$
$16^{\frac{1}{2}}-7{\frac{1}{2}}=4-\sqrt 7=1.35425\dots $
