How do you evaluate the definite integral #int (2x+1)/(x+1)# from #[0,e-1]#?

Question

Aurora Alvarado · Accepted Answer

#2e - 3#

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!

Charles Arocha · Answer

undefined To evaluate the definite integral ∫(2x + 1)/(x + 1) from 0 to e - 1, you can first perform polynomial long division to rewrite the integrand as 2 - 1/(x + 1). Then integrate each term separately. The integral of 2 with respect to x from 0 to e - 1 is 2(e - 1). The integral of 1/(x + 1) from 0 to e - 1 can be found using the natural logarithm, resulting in ln(e) - ln(1) = 1 - 0 = 1. Therefore, the value of the definite integral is 2(e - 1) + 1.