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How do you integrate #(2x)/(x²-1) #?

1 Answer

May 28, 2016

#int (2x)/(x^2-1) d x=l n(x^2-1)+C#

Explanation:

#int (2x)/(x^2-1) d x=?#

#x^2-1=u#

#2x d x=d u#

#int (2x)/(x^2-1) d x=int (d u)/u=l n u +C#

#int (2x)/(x^2-1) d x=l n(x^2-1)+C#

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