Integration and derivative together I have the  function $f(x)=\ln(x+\sqrt{x^2+1})$ and I have already find that $$f'(x)=\frac{1}{\sqrt{x^2+1}}$$ and I want to find the $$\int_0^1{\sqrt{x^2+1}}dx$$ so:
$$\int_0^1{\sqrt{x^2+1}}dx=\int_0^1{\frac{1}{f'(x)}}dx$$
but I am stack and I don't know how to continue. Any ideas?

Do an integration by parts. Let $I = \int \sqrt{x^2 + 1}\;dx$. Then
\begin{align}
I &= x\sqrt{x^2+1} - \int\frac{x^2}{\sqrt{x^2+1}}\;dx \\
&= x\sqrt{x^2+1} - \int\frac{x^2+1-1}{\sqrt{x^2+1}}\;dx \\
&= x\sqrt{x^2+1} - I + \int\frac{1}{\sqrt{x^2+1}}\;dx
\end{align}
So the integral $I$ is:
$$I = \frac{1}{2}\left(x\sqrt{x^2+1} + \log(x + \sqrt{x^2+1})\right)$$
Now, just plug in your integration limits!
EDIT: If you plug in your limits, you should get:
$$\int_0^1\sqrt{x^2+1}\;dx = \frac{1}{2}\left(\sqrt{2} + \log(1+\sqrt{2})\right)$$
