How would you solve the intigral from -1 to 2 for the equation (x/(x^2+1))dx?

Question

bahrom7893 · Answer

let u =x^2 +1
du = 2xdx
rewrite the integral as:
1/2 Integral (2xdx/(x^2+1))

bahrom7893 · Answer

1/2 Integral (du/u) = 1/2 Ln|u| = 1/2 Ln(x^2+1) + C

bahrom7893 · Answer

Note you dont need abs value any more because x^2 is always positive and positive plus one is still positive!

bahrom7893 · Answer

Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)

anonymous · Answer

TYVM

anonymous · Answer

wait...
 is the anser 1/2ln5/2 or 1/2ln3?

bahrom7893 · Answer

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 · Answer

oh wait no

bahrom7893 · Answer

1/2ln5/2

bahrom7893 · Answer

2^2 is 4, not 2...
(1/2){Ln(4+1) - Ln(1+1)} = (1/2) {Ln5 - Ln2} = (1/2)Ln(5/2)

anonymous · Answer

yep ok that helps alot ty