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

1 Answer
Oct 14, 2016

#int_0^(e-1)(x+2)/(x+1)dx=e#

Explanation:

#int_0^(e-1)(x+2)/(x+1)dx#

Either rewrite the function using the following procedure or long divide #(x+2)/(x+1)# for the same results:

#=int_0^(e-1)(x+1)/(x+1)+1/(x+1)dx#

#=int_0^(e-1)1+1/(x+1)dx#

Both of these have common antiderivatives:

#=[x+ln(abs(x+1))]_0^(e-1)#

#=[e-1+ln(abs(e-1+1))]-[0+ln(abs(0+1))]#

#=(e-1+ln(e))-(ln(1))#

#=(e-1+1)-0#

#=e#