Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

TYVM

Answer 6

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

Answer 7

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)

Answer 8

oh wait no

Answer 9

1/2ln5/2

Answer 10

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

Answer 11

yep ok that helps alot ty