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

Question

Christopher Agresta · Accepted Answer

#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#

Paisley Johnson · Answer

undefined To evaluate the definite integral $\int_0^{e-1} \frac{x+2}{x+1} \, dx$, we can use the technique of integration by substitution. Let $u = x + 1$, then $du = dx$.

Now, when $x = 0$, $u = 0 + 1 = 1$, and when $x = e - 1$, $u = (e - 1) + 1 = e$.

The integral becomes: 
$$\int_1^e \frac{u + 1}{u} \, du$$

This simplifies to:
$$\int_1^e \left(1 + \frac{1}{u}\right) \, du$$

Now, integrate term by term:
$$= \left[u + \ln|u|\right]_1^e$$

Substitute back $u = x + 1$:
$$= \left[(x + 1) + \ln|x + 1|\right]_0^{e-1}$$

$$= ((e - 1) + \ln|e - 1|) - (0 + \ln|0 + 1|)$$

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

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