First, check to see if the integral is continuous on #[a, b]#. Since this is a rational function, we check for vertical asymptotes. When #x= -1#, the denominator equals 0 and therefore the equation of the asymptote is #x= -1#. Since #-1# is not in #[0, e - 1]#, we can say that this is a definite integral.
This would be integrated using partial fractions.
#A/1+ B/(x + 1) = (2x+ 1)/(x+ 1)#
#A(x + 1) + B = 2x + 1#
#Ax + A + B = 2x + 1#
We can now write a system of equations:
#{(A = 2), (A + B = 1):}#
Solving, we have #A = 2# and #B = -1#.
The integral becomes:
#int_0^(e- 1) 2 - 1/(x + 1)dx = int_0^(e - 1) 2dx - int_0^(e - 1) 1/(x + 1)dx#
These two integrals can be readily integrated.
#=[2x]_0^(e - 1) - [ln(x + 1)]_0^(e - 1)#
Evaluate using the second fundamental theorem of calculus, which states that
#=2(e - 1) - 2(0) - (ln(e - 1 + 1) - ln(0 + 1))#
#=2e - 2 - ln(e) + ln(1)#
#= 2e - 2 - 1 + 0#
#=2e - 3#
Hopefully this helps!
