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

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Answer 1 · 2016-10-14T23:26:22

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

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

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:

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

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

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

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

$=e$

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