TITLE: On the evaluation of the integral $\int_{0}^{1}\sqrt{x^2+1}\,dx$
QUESTION [7 upvotes]: The integral $\displaystyle \int_{0}^{1}\sqrt{x^2+1}\,dx$ can be evaluated with the standard technic of sub $u=\tan \theta$. However, in the book says evaluate the integral without trigonometric substitution.
I can't find a way. I applied the sub $u=1-x$ and it got a little messie.
I also tried to approach it geometrically, but again I had a problem because the graph is not a common shape thus saying that the integral is equal to the area of that shape.
For example if I had the integral $\displaystyle \int_0^1 \sqrt{1-x^2}\,dx$ then this equal to aread of the quadrateral etc. In the case of the other integral I don't see something like this.
If someone could give me a hint,that would be nice!!

REPLY [12 votes]: Hint. Substitute: $\sqrt{x^2+1}=x+t$. This transforms your algebraic integral into a purely rational one:
$$\begin{align}
\int_{0}^{1}\sqrt{x^2+1}\,\mathrm{d}x
&=\int_{1}^{\sqrt{2}-1}\frac{t+\frac{1}{t}}{2}\cdot\frac{(-1)(t^2+1)}{2t^2}\,\mathrm{d}t
\end{align}$$