Evaluate
integral sign (x-1/(2x))^2 dx=
(x - 1/(2x))^2 = x^2 - 1 + 1/(4x)
((x-1)/(2x))^2 = (x^2 - 2x + 1)/(4x^2) = 1/4 - 1/(2x) + 1/(4x^2)
Not sure which you meant, but either one is just a sum of terms, which are just powers of x. Just use the normal power rule (or lnx for 1/x) and things are easy.
oops. Top one should end with 1/(4x^2)
Thanks
To evaluate the given integral:
Step 1: Expand the expression inside the integral
(x - 1/(2x))^2 = (x^2 - 2x*(1/(2x)) + (1/(2x))^2
= x^2 - 1 + (1/(4x^2))
Step 2: Rewrite the integral
∫ (x^2 - 1 + (1/(4x^2))) dx
Step 3: Evaluate each term separately
∫ x^2 dx - ∫ dx + ∫ (1/(4x^2)) dx
Step 4: Evaluate each term using integration rules
Integral of x^2 dx = (1/3)x^3 + C
Integral of dx = x + C
Integral of (1/(4x^2)) dx = (-1/(4x)) + C
Step 5: Substitute back the evaluated terms into the original expression
(1/3)x^3 - x - 1/(4x) + C
Therefore, the evaluated integral is (1/3)x^3 - x - 1/(4x) + C.