`int_0^1 sqrt(x^2 + 1) dx` Evaluate the integral

`int_0^1sqrt(x^2+1)dx`

Let's first evaluate the definite integral using the standard integral,

`intsqrt(a^2+x^2)dx=(xsqrt(a^2+x^2))/2+a^2/2ln|x+sqrt(a^2+x^2)|+C`

`intsqrt(x^2+1)dx=(xsqrt(1^2+x^2))/2+1^2/2ln|x+sqrt(1^2+x^2)|+C` 

`=(xsqrt(1+x^2))/2+1/2ln|x+sqrt(1+x^2)|+C`

Now let's evaluate the definite integral,

`int_0^1sqrt(x^2+1)dx=[(xsqrt(1+x^2))/2+1/2ln|x+sqrt(1+x^2)|]_0^1`

`=[(1sqrt(1+1))/2+1/2ln|1+sqrt(1+1)|]-[(0sqrt(1+0))/2+1/2ln|0+sqrt(1+0)|]`

`=[sqrt(2)/2+1/2ln|1+sqrt(2)|]-[0+1/2ln(1)]`

`=sqrt(2)/2+1/2ln(1+sqrt(2))`

Since `arcsinh(x)=ln(x+sqrt(x^2+1))`

`=1/2(sqrt(2)+arcsinh(1))`