- #1
ruiwp13
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Homework Statement
Solve the following integral:
∫arctg(1/√x)dx
Homework Equations
∫u'/1+u^2 = arctg(u)+c
∫u.v' = u.v - ∫ u'.v
The Attempt at a Solution
So , I tried to define u as arctg(1/√x) but I'm having trouble finding the du. The derivate of arctg(1/√x) is -1/2*x^(3/2), so it stays -1/2*x^(3/2)/1-(1/√x)^2 (converting to the derivate u'/1+u^2) ? I was thinking about getting the derivate of arctg(1/√x) and doing this by parts. dv = dx , u = arctg(1/√x)