anonymous
  • anonymous
How would you solve the intigral from -1 to 2 for the equation (x/(x^2+1))dx?
Mathematics
chestercat
  • chestercat
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bahrom7893
  • bahrom7893
let u =x^2 +1 du = 2xdx rewrite the integral as: 1/2 Integral (2xdx/(x^2+1))
bahrom7893
  • bahrom7893
1/2 Integral (du/u) = 1/2 Ln|u| = 1/2 Ln(x^2+1) + C
bahrom7893
  • bahrom7893
Note you dont need abs value any more because x^2 is always positive and positive plus one is still positive!

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bahrom7893
  • bahrom7893
Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)
anonymous
  • anonymous
TYVM
anonymous
  • anonymous
wait... is the anser 1/2ln5/2 or 1/2ln3?
bahrom7893
  • bahrom7893
oh sorry forgot to evaluate: 1/2 Ln(x^2+1) from -1 to 2: (1/2){Ln(2+1) - Ln(1+1)} = (1/2) {Ln3 - Ln2} = (1/2)Ln(3/2)
bahrom7893
  • bahrom7893
oh wait no
bahrom7893
  • bahrom7893
1/2ln5/2
bahrom7893
  • bahrom7893
2^2 is 4, not 2... (1/2){Ln(4+1) - Ln(1+1)} = (1/2) {Ln5 - Ln2} = (1/2)Ln(5/2)
anonymous
  • anonymous
yep ok that helps alot ty

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