Prove that $\int^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$ 
Prove that $\int\limits^{1}_{0} \sqrt{x^2+x}\,\mathrm{d}x < 1$

I'm guessing it would not be too difficult to solve by just calculating the integral, but I'm wondering  if there is any other way to prove this, like comparing it with an easy-to-calculate integral. I tried comparing it with $\displaystyle\int\limits^{1}_{0} \sqrt{x^2+1}\,\mathrm{d}x$, but this greater than $1$, so I'm all out of ideas.
 A: HINT:
$x^2+x<x^2+x+\frac14=\left(x+\frac12\right)^2$
A: A method which is generally applicable to concave functions:
If $f:[a, b] \to \Bbb R$ is concave then its graph lies below any tangent line:
$$
 f(x) \le f(c) + f'(c)(x-c) \, .
$$
If one chooses $c =(a+b)/2$ then the integral of the second term on the right over the interval $[a, b]$ is zero, and therefore
$$
 \int_a^b f(x) \, dx \le (b-a) \cdot f\left( \frac{a+b}{2}\right) \, .
$$
In our case this gives
$$
 \int_0^1 f(x) \, dx \le f\left(\frac 12 \right) = \frac 12 \sqrt 3 < 1 \, .
$$
The estimate $\frac 12 \sqrt 3 \approx 0.86603$ comes pretty close to the exact value of the integral, which is $\approx 0.84032$.
A: Well, you could use the AM-GM inequality:
$$\sqrt{x^2 + x} = \sqrt{x(x+1)} < \frac{x + (x+1)}{2} = x + \frac{1}{2}$$
and then
$$\int_0^1 \sqrt{x^2 + x} \, dx < \int_0^1 x + \frac{1}{2} \, dx = \frac{1}{2} + \frac{1}{2} = 1.$$
A: Yet another way: If $0 < x < 1$, then $x^2 < x$. So
$$\int_0^1 \sqrt{x^2+x} \ dx < \int_0^1 \sqrt{2x}\ dx = \frac{2 \sqrt{2}}{3} < 1$$
(This bound is about 0.9428, so not as good as Martin R's answer)
A: Still simpler.
For $0<x<1$ we have $x^2<x$.
Hence $\sqrt{x^2+x}\lt \sqrt{x+x}=\sqrt{2}\sqrt{x}$, and the integral can be estimated as $\int_0^1 \sqrt{x^2+x}<\sqrt{2}\int_0^1 \sqrt{x}=\frac{2}{3}\sqrt{2}=\sqrt{\frac{8}{9}}\lt 1$.
QED.
