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## Homework Statement

∫ 1/(x√(25-x^2)) dx

## Homework Equations

by trigonometic substitution I got the following equations

x = 5 sin θ

dx = 5 cos θ dθ

√(25-x^2) = 5 cos θ

θ = sin^-1 (x/5)

## The Attempt at a Solution

I solved it and got the following (so far, it is okay with wolfram alpha):

∫ 1/(x√(25-x^2)) dx = (1/5) ∫ csc θ dθ

according to me, the result should be (1/5) ( ln ( 5/x - √(25-x^2)/x ) )

but according to wolfram alpha it should be (1/5) ( ln (x) - ln ( √(25-x^2) + 5 ) )

are these two results equivalent?