I have met this kind of problem today.in fact, I spent hours trying to solve this problem on its own. My method was just to calculate the indefinite integral. I looked at Wolfram Alpha after I failed. Wolfram couldn't evaluate this integral. I don't know the spesific reason.
The integral:
$$\int_{0}^{16}\frac{dx}{\sqrt{x^2+9}-\sqrt{x}}$$
On
As observed in the comments, this would be much easier if $\sqrt{x}$ were replaced with $x$.
Nevertheless, we can still evaluate this integral by hand by rationalizing the denominator and splitting the integrand into two summands, giving: $$\int \frac{\sqrt{x}}{x^2 - x + 9} dx + \int \frac{\sqrt{x^2 + 9}}{x^2 - x + 9} dx .$$ To evaluate the first integral, we can rationalize by reverse-substituting $x = u^2, dx = 2 u \,du$, giving $$2 \int \frac{u^2}{u^4 - u^2 + 9}.$$ To rationalize the second integral, apply the Euler substitution $\sqrt{x^2 + 9} = x + t$ (or $x = \frac{9 - t^2}{2 t}$, $dx = -\frac{t^2 + 9}{t^2}$), giving $$-\int \frac{(u^2 + 1)^2 \,du}{u (u^4 + 2 u^3 + 18 u^2 - 18 u + 81)} .$$ We've now reduced the problem to evaluating two integrals of rational functions, so we can in principle apply the method of partial fractions. Generically the quartics appearing in the denominators might mean an awful mess, but those appearing here both factor into quadratics over $\Bbb Q[\sqrt{7}]$ (this just means that we can write the real partial fractions decomposition using only rational numbers and $\sqrt{7}$.) The radicals in the coefficients make this factorization and the subsequent integration ugly, but as soon as we have the quadratic factors of the quartics, the rest of the problem is procedurally routine.
The antiderivative is long and unenlightening, so I won't reproduce here, but Maple finds for the definite integral the closed-form expression $$\tiny-\frac{1}{\sqrt{5}}\arctan \left( 5\,{\frac {-221\,\sqrt {53}+371\,\sqrt { 5}}{91\,\sqrt {5}\sqrt {53}+22525}} \right) +\frac{\pi}{\sqrt{5}} -\frac{1}{\sqrt{7}}\ln \left( {\frac {1}{166}}\,\sqrt {265}\sqrt {7}+{\frac {95 }{498}}\,\sqrt {7}+{\frac {35}{498}}\,\sqrt {265}+{\frac {19}{166}} \right) +\ln \left( \frac{3}{-16+\sqrt {265}} \right) , $$ which should agree with the other answers here after some simplification.
According to Maple 18, the integral is equal to \begin{align*} &\frac15\arctan \left( \frac{\sqrt {5}}{5} \right) \sqrt {5}+\frac{1}{14}\sqrt {7} \ln \left( \sqrt {7}+1 \right) -\frac{1}{14}\sqrt {7}\ln \left( \sqrt {7}-1 \right) \\ &-\frac{1}{14}\sqrt {7}\ln \left( \sqrt {5}\sqrt {7}\sqrt {53}+19 \right) +\frac{1}{14}\sqrt {7}\ln \left( \sqrt {5}\sqrt {7}\sqrt {53}-19 \right) \\ &+\frac{1}{5}\sqrt {5}\arctan \left( {\frac {13\,\sqrt {53}}{265}} \right) +\frac{1}{5}\sqrt {5}\arctan \left( \frac85\sqrt {5}-\frac{1}{5}\sqrt {7}\sqrt {5} \right) \\ &+\frac{1}{5}\sqrt {5}\arctan \left( \frac85\sqrt {5}+\frac{1}{5} \sqrt {7}\sqrt {5} \right)+\frac{1}{14}\sqrt {7}\ln \left( 19-4\,\sqrt {7} \right) \\ &-\frac{1}{14}\sqrt {7}\ln \left( 19+4\,\sqrt {7} \right) +\ln \left( 3 \right) -\ln \left( -16+\sqrt {5}\sqrt {53} \right). \end{align*} Enjoy.