How do you evaluate the definite integral #int (x+2)/(x+1)# from #[0, e-1]#?
1 Answer
Oct 14, 2016
Explanation:
#int_0^(e-1)(x+2)/(x+1)dx#
Either rewrite the function using the following procedure or long divide
#=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#
