To start with, we must find the antiderivative of the integrand, #1/(5x+1)^2#. We do this using a u-substitution. #int1/(5x+1)^2#, #u=5x+1# and #du=5dx#. We don't see #5dx# in the integrand but we can rewrite divide both sides by 5 to get #1/5du=dx# and then we substitute and integrate #1/5int1/u^2du# which results in #-1/(5u)# and then we plug u in to get #-1/(5(5x+1))#.
Now, we must evaluate the antiderivative at both bound values. We start with the top bound, #x=oo#. Plug in, and we get #-1/(5(5(oo)+1))#. The five multiplied by infinity results in infinity and then add 1 and that is still infinity and when you multiply that infinity by 5, it still results in infinity. So all that simplifies to #-1/oo#. Since the denominator is infinitely big, the overall value is infinitely small and that result is equivalent to #0#.
The next value we calculate is at the next bound, #x=1#. Plug in and it results in #-1/26#. Lastly, we take the first value and subtract the second value from it. This looks like #0-(-1/26)# and those two negatives cancel and leave us with #1/26#
