The integral problem: $\int_{0}^{16}\frac{dx}{\sqrt{x^2+9}-\sqrt{x}}$

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:

$$\int_{0}^{16}\frac{dx}{\sqrt{x^2+9}-\sqrt{x}}$$ • Rationalize it and seperate... – Szeto Aug 16 '18 at 11:35
• @Szeto That was my first thought too, but unless I'm missing something, it doesn't work. You'd really need $\sqrt{x^2 + 9} - x$ on the denomintor instead. – Theo Bendit Aug 16 '18 at 11:36
• @Theo Bendit You are right. – Math Aug 16 '18 at 11:37
• The inverse symbolic calculator's interpretation of Wolfram Alpha's numerical result is $-\frac{3}{2}-\frac{9\pi}{7}+\frac{47e}{14}$. The $e$ dependence looks implausible, which is unfortunate because I was hoping it would give us a clue. – J.G. Aug 16 '18 at 11:39
• @TheoBendit See Travis’s answer. That’s what I mean. – Szeto Aug 16 '18 at 12:49

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.

• That’s exactly what I meant. – Szeto Aug 16 '18 at 12:46
• (+1) Thank you for answer. But if this integral indefinite , is it possible to find a closed-form expression? – Math Aug 16 '18 at 13:06
• @Math It will be, yes. tedious, but definitely possible. – J.G. Aug 16 '18 at 14:18

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.

• :):):):):):) (+1) – Math Aug 16 '18 at 11:45
• Did you see J.G.'s comment? – Math Aug 16 '18 at 11:47
• I did, but this wasn't inverse symbolically calculated. This was Maple working it out analytically. It took longer than most integral computations too. – Theo Bendit Aug 16 '18 at 11:50
• @Zacky If you can fit this answer into a comment, be my guest! – Theo Bendit Aug 16 '18 at 11:51
• @Math This answer is much more plausible than the one I got from the ISC. If someone with maple 16 access change the $16$ to several other values, they can probably spot an antiderivative, then prove it's right just by differeniating. – J.G. Aug 16 '18 at 11:55