The answer is A. 5 is not an even number, so both events can't take place together.
The average speed is given by:
average speed=(total distance)/(total time)
total time for the first 30 miles was:
time=distance/speed
=30/30
=1 hr
total time for the remaining 30 miles
=30/60
=0.5
hence the average speed will be:
speed=(30+30)/(1+0.5)=60/1.5=40 miles per hour
If the integral is simply

then notice that

which means you can compute the integral easily with a substitution

Under this transformation, the integral is

On the other hand, in case you're missing a symbol and the integral is actually

then first carry out the division:

Now,
, so to integrate the remainder term you can decompose it into partial fractions:





Then the integral would be

which can be rewritten in several ways, such as
