Integration
asked Sep 13, 2017 in Other Math Topics by Iviwe

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1) Let I=∫(x+1)dx/√(1-x^2)=xdx/√(1-x^2)+dx/√(1-x^2)+C.

Let u^2=1-x^2, 2udu/dx=-2x, so udu=-xdx; let x=sin(p), so dx=cos(p)dp, dp=dx/√(1-x^2).

I=-∫du+∫dp=-u+p=-√(1-x^2)+arcsin(x)+C.

2) x^2-2x+5 can be written: x^2-2x+1+4=(x-1)^2+2^2.

Let x-1=2tanp, then (x-1)^2+4=4(tanp)^2+4=4(secp)^2; dx=2(secp)^2dp.

Also tanp=(x-1)/2 and p=arctan((x-1)/2).

∫dx/(x^2-2x+1)=∫2(secp)^2dp/4(secp)^2=½∫dp=p/2+C=½arctan((x-1)/2)+C.

answered Sep 13, 2017 by Rod Top Rated User (537,640 points)

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