# Math

Calculate the integrals by partial fractions and using the indicated substitution. Show the results you get are the same.

dx/1-x^2; substitution x= sin pheta

I understand how to do the partial fraction part, but not the second part and I don't know how they are similar. Any help would be appreciated on what to do

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1. They want you to do it two ways. The first is to change it to
dx/[2(1+x)] + dx/[2(1-x)]
which integrates to
(1/2)[ln(1+x) - ln(1-x)]
= (1/2)ln[(1+x)/(1-x)]

In the substitution method, with x = sin u
dx = cos u du

Integral dx/(1-x^2)= cos u du/1-sin^2u
= Integral du/cos u = Integral (sec u)
= (1/2)log[(1+sinu)/(1-sinu)]
= (1/2)log[(1+x)/(1-x)]

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posted by drwls

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