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

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Answer 1 · 2018-06-24T12:20:57

$6.368995013$

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
$approx 6.368995013$

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