積分題目:已知 y=√(〖2x〗^2+1)。若 f(x)=2 dy/dx，求 ∫2 0 f(x) dx 的值。?

Sol
y=[(2x)^2+1]^(1/2)
y^2=4x^2+1
2ydy=8xdx
dy/dx=8x/(2y)=4x/y
f(x)=2dy/dx=8x/y=8x/[(2x)^2+1]^(1/2)
A=∫(0 to 2)_f(x)dx
=∫(0 to 2)_ 8x/[(2x)^2+1]^(1/2)dx
Set p=2x
dp=2dx
A=∫(0 to 2)_ 2*(2x)/[(2x)^2+1]^(1/2)d(2x)
=∫(0 to 1)_ 2*p/(p^2+1)^(1/2)dp
=∫(0 to 1)_ (p^2+1)^(－1/2)dp^2
=2(p^2+1)^(1/2)｜(0 to 1)
=2√2－2