Integration by trigonometric substitution

[cesaruelas]

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?

 

[LCKurtz]

Differentiate the answers and see if you get the integrand. That will tell you.

 

[SammyS]

\displaystyle \ln (x) - \ln ( \sqrt{25-x^2} + 5 )=\ln\left(\frac{x}{\sqrt{25-x^2} + 5}\right)

\displaystyle =-\ln\left(\frac{\sqrt{25-x^2}}{x}+\frac{5}{x}\right)​

So, there's only a disagreement about the sign.