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integrate x^2/(x^2+1)^2
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i would do partial fractions
x^2 / ( (x^2 + 1) ( x^2 + 1)) , the denominator are irreducible quadratics
x^2 / ( (x^2 + 1) ( x^2 + 1)) = ( A x + b ) / (x^2 + 1) + (Cx + D ) / ( X^2 + 1)^2
x^2 = (Ax+b)(x^2 + 1) + Cx + d x^2 = Ax^2 + bX^2 + Ax + B + CX + D x^2 = Ax^3 + Bx^2 + (A + C) x + B + D so 1x^2 = Bx^2 so B=1
A = 0 , B=1 , D= -1 , C= 0
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so we have intergral x^2/(x^2+1)^2 = int 1/(x^2 +1) - integral 1 / (x^2 +1)^2 the first integral is an arctan, the second one is ... let methink
hmmm, i hope this isnt circular
nevermind, its faster if you do trig substitution
this freaking website just deleted what i wrote.
very frustrating!!!!
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