As the present time is 2:14 AM, I will get you started with the first problem, and return in a few hours to finish up (hopefully).
First, we interpret the improper integral, i.e., int[f(x) dx] ; [a,inf], as the limit of the definite integral on interval [a,n] as n --> inf. In the present case the lower limit of integration is 3, making the interval [3,n].
Now, if we complete the square on denominator x^2+2x+10, the result is (x^2+2x+1) + 9, or (x+1)^2 + 9. So we have:
Lim_n->inf [12 int (dx / ((x+1)^2 + 9))]. Yes, I know it's time to learn latex.
Letting x+1 = 3tan(ß) gives dx = 3sec^2(ß) * dß. Hence the integral becomes:
12*3 int [sec^2(ß) * dß / (9tan^2(ß) + 9)] = 4 int [sec^2(ß) * dß / (tan^2(ß) + 1)].
Applying the identity tan^2(ß) + 1 = sec^2(ß) leaves 4 int [sec^2(ß) * dß / sec^2(ß) ] = 4 int (dß) = 4ß.
Now, recall the substitution x+1 = 3tan(ß). Because the interval on x is [3,n], we can derive an equivalent interval on ß by solving for ß in,
1) 3+1 = 3tan(ß) and 2) n+1 = 3tan(ß). Thus the new interval is [arctan(4/3), arctan((n+1)/3)]. And the integral, i.e., 4ß ; [x=(3,n)] is evaluated as:
4ß ; [arctan(4/3), arctan((n+1)/3)] = 4 * [arctan((n+1)/3) - arctan(4/3)].
I leave the finishing details to you. I'm off to catch up on some sleep.