Need some help finding the definite integral of x/(1+x^2) on [10,infinity]?

Question

Please explain how to get the answer if you can. Thanks in advance.

Answer 1

You first have to find the anti-derivative. If you use the substitution u=1+x^2, you will find an anti-derivative to be
(1/2)ln(1+x^2).
This is evaluated at 10 and at a number b which will go to infinity. But as b-->infty, 1+b^2-->infty, so ln(1+b^2)-->infty. Thus, the integral diverges.

Answer 2

The answer is infinity.

You have to set it up from 10 to B and then take the limit as B goes to infinity after you find the definite integral from 10 to B.
The integral of x/(1+x^2) simplifies to 1/2ln(1+x2) using u-substitution. The definite integral simplifies to 1/2[ln(1+B^2)-ln(1+10^2)]. The limit of the definite integral simplifies to infinity. Thus the answer to the definite integral is infinity.

Answer 3

First, compute ∫(10 to b) x/(1+x^2) dx
Substitute u = 1 + x^2, du = 2x dx
= ∫ (101 to 1+b^2) (du / (2u))
= (1/2) [ln u] [101 to 1+b^2]
= (ln (1+b^2) - ln 101) / 2
Now as b -> ∞, this integral also goes to ∞. So the (infinite) integral is undefined.