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\[\huge \int\limits_{}^{}\frac{ 3 }{ (2-x)^{2} }dx\]
Pull the 3 out, u = 2-x, du = -dx you'll get 1/u^2du ..etc.etc, gtg now, take care and good night! :)
@satellite73
@sourwing
\[u=2-x,du=-dx\] should get it in one step
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Yes i got that
\[-3\int \frac{du}{u^2}\] etc
after that
\[\frac{1}{u^2}=u^{-2}\]power rule backwards
|dw:1399611310736:dw|
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