# Calculus

posted by .

Evaluate
integral sign (x-1/(2x))^2 dx=

• Calculus -

(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.

• Calculus oops -

oops. Top one should end with 1/(4x^2)

• Calculus -

Thanks

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