Question

What is the arc length of `f(x)=(2x^2–ln(1/x+1))` on `x in [1,2]`?

Answer 1

`6.368995013`

Explanation:

Given
`f(x)=2x^2+ln(1/x+1)`
so we get

`f'(x)=4x+1/(x^2(1/x+1))`
this is

`f'(x)=4x+1/(x+x^2)`

`f'(x)=(4x(x+x^2)+1)/(x+x^2)`

`f'(x)=(4x^3+4x^2+1)/(x+x^2)`
and our integral will be

`int_1^2sqrt(1+((4x^3+4x^2+1)/(x(x+1)))^2)dx`
By a numerical method we get
`approx 6.368995013`