Question regarding an integral that does not converge and and why the same one does! Please help!!!

The problem I am working with is the following:

The integral from [-infinity, infinity] of (1+x) / (1+ (x^2)) dx.

I am wondering how this integral can have a solution that shows it does not converge, and at the same time a solution of pi. Here's my solution for the first solution:

ln |1 + infinity squared| + (1/2)ln |1+ infinity squared| - ln |1 + (-infinity squared)| - (1/2)ln |1+ (-infinity squared)|

Is this right? So the answer for why it does not converge is because the solution goes to infinity?
And I'm not sure where to start to prove that this same integral goes to pi. Please help me/guide me through!!!

your integration is not correct. lets forget about the limits to start

∫ (1+x)/(1 + x^2) dx

write this as two cases to consider,

1 / (1 + x^2) + x/(1 + x^2) dx

first integral is arctan(x)

second one,
u = 1 + x^2, du = 2x dx
then you get 1/2 log( 1 +x^2 )

so our integral is,

1/2 log( 1 + x^2 ) + arctan(x)

now for the limits, if we say the integral is from a to b, where lim a→-∞, lim b→∞, we have

lim a→-∞, lim b→∞ 1/2 log( 1 + b^2 ) + arctan(b) - 1/2 log( 1 + a^2 ) - arctan(a)

lets do some clean up first:

lim a→-∞, lim b→∞ 1/2 log( (1 + b^2) / (1 + a^2) ) + arctan(b) - arctan(a)

clearly this imit is undefined.

IF you make the mistake and say the integral goes from -a to a, where lim a-> ∞, then you get:

lim a→∞ 1/2 log( 1 + a^2 ) + arctan(a) - 1/2 log( 1 + a^2 ) - arctan(-a)
lim a→∞ arctan(a) - arctan(-a)
lim a→∞ 2arctan(a) = π

but that is VERY wrong because you should not assume that -∞ and ∞ are the same as -2 and 2.

-∞ + ∞ DOES NOT EQUAL 0

First of all, you didn't take the integral correctly.

∫ (1+x) / (x²+1) dx

(dx / x² + 1 + ∫x / x² + 1

arctan(x) +(ln(x²+1)) / 2

Should be your integral