Solving $\int_0^1 \sqrt{x^2+1}$ with Euler substitution So I was trying to solve this integral 
$$\int_0^1 \sqrt{x^2+1} \, dx$$ 
in two different ways with Euler substitution. So: 


*

*$(x^2+1)=x+t$, so that $x=(1-t^2)/(2t)$,

*$(x^2+1)=x*t+1$, $x=2t/(1-t^2)$.


But I am not sure how to proceed, and how do the integral and the limits change... Any tips? Thanks :)
 A: Put $x=\sinh (t)=\frac {e^t-e^{-t}}{2} $
then
$x=0$ gives $t=0$
$x=1$ gives $t $ such that 
$$e^t-e^{-t}=2$$
that is
$$(e^t)^2-2e^t-1=0$$
$$t=\ln (1+\sqrt {2})$$
thus the integral becomes
$$\int_0^{\ln (1+\sqrt {2})}\cosh^2 (t)dt $$
$$=\int_0^{\ln (1+\sqrt {2})}\frac {1+\cosh (2t)}{2}dt $$
$$=\frac {1}{2}\ln (1+\sqrt {2})+\frac {1}{8}((1+\sqrt {2})^2-(1-\sqrt {2})^2) $$
i am sure you can finish.
A: @Salahamam With Euler substitution it is meant a particular kind of substitution  discovered by Euler for integrals containing a squareroot and a quadratic polynomial inside. See my link in a comment. It is not about $e^{it}$ here. The substitution in question is $\sqrt{x^2+1}=x+t$. Now solving this for $x$ to get: $x=\frac{1-t^2}{2t}$ and so $dx=\frac{-2t^2-2}{4t^2}dt$. The given integral is of the form $\int\frac{1-t^2+2t^2}{2t}*\frac{-2t^2-2}{4t^2}dt$. Enfin, this is now all polynomial work where the denominator is a monomial and so after the numerator is worked out, one can split fractions and integrate. Please give it a try from here. This is just (annoying) algebra...
