Posted by Jessica on Monday, February 25, 2008 at 3:52am.
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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