How do you evaluate the integral #int 1/(xsqrt(4-x^2))#?

1 Answer
Noah G
Oct 21, 2017

#I = 1/4ln|sqrt(4 - x^2) - 2| - 1/4ln|2 + sqrt(4 -x^2)| + C#

Explanation:

From first glance we would think this integral requires trig substitution but in fact it doesn't. If we let #u = sqrt(4 - x^2)#, then #du = (-2x)/(2sqrt(4 - x^2)) dx# and so #dx = du(sqrt(4 - x^2))/(-x)#

Substituting, we get:

#I = int 1/(xsqrt(4 - x^2)) * sqrt(4 - x^2)/(-x)du#

#I = int -1/x^2du#

But since our bounds of integration are in #u#, this won't quite work. But notice that #u^2 = 4 - x^2# and #-x^2 = u^2 - 4#.

#I = int 1/(u^2 - 4) du#

We can now use partial fractions to solve.

#A/(u - 2) + B/(u + 2) = 1/((2 - u)(2 + u))#

#A(2 + u) + B(u - 2) = 1#

#2A + Au -2B + Bu = 1#

#(2A - 2B) + (Au +Bu) = 1#

We can now write a system of equations.

#{(2A - 2B = 1), (A + B = 0):}#

Solving for #A# and #B#, we get

#4B = 1#

#B = -1/4#

So #A = 1/4#. The integral now becomes

#I = int 1/(4(u - 2)) - 1/(4(2 + u)) du#

This can be easily integrated now.

#I = 1/4ln|u - 2| - 1/4ln|2 + u| + C#

Reverse our substitutions now.

#I = 1/4ln|sqrt(4 - x^2) - 2| - 1/4ln|2 + sqrt(4 -x^2)| + C#

Hopefully this helps!

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