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November 28, 2014

November 28, 2014

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

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

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

- Math -
**drwls**, Monday, February 25, 2008 at 7:20amThey 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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