Show there is a unique solution to the equation F(x)=c if c>0 Define $F(x)=\int_1^x \frac{1}{2\sqrt{t} -1} dt$ for $x\ge1$. If c>0, prove there is a unique solution to the equation F(x)=c, for x>1.
I know I need to use the intermediate value theorem, but how do I get there? 
 A: The function $F(x)$ is continuous, and $F(1)=0$. 
Also, $F'(x)=\frac{1}{2\sqrt{x}-1}\gt 0$, so $F(x)$ is increasing. 
Note that
$$F(x)\gt \int_1^x\frac{1}{2\sqrt{t}}\,dt=\sqrt{x}-1.$$
Thus $F(x)\to\infty$ as $x\to\infty$. In particular there is a $b$ such that $F(b)\gt c$.
Now for existence we can use the Intermediate Value Theorem. Let $a=1$. We have $F(a)\lt c$ and $F(b)\gt c$. so there is an $x$ between $a$ and $b$ such that $F(x)=c$. 
For uniqueness, use the fact that $F(x)$ is increasing on the interval $[1,\infty)$.  
A: From the fundamental theorem of calculus, we have $$F'(x)=\frac{1}{2\sqrt{x} -1}$$ which is positive and so $F(x)$ is increasing. As written by André Nicolas, $F(x)$ goes to infinity. So $F(x)=c$ has a unique solution for $x \gt 1$. 
We could even go further and compute the integral to get $$F(x)=\sqrt{x}+\frac{1}{2} \log \left(2 \sqrt{x}-1\right)-1$$ and the solution of $F(x)=c$ is given by $$x_{sol}=\frac{1}{4} \left(1+W\left(e^{2 c+1}\right)\right)^2$$ where appears the Lambert function.
