Question

What is the antiderivative of `(x+2)/(x+1)`?

Answer 1

`x+lnabs(x+1)+C`

Explanation:

Using substitution:

Set `u=x+1`. This means that `x=u-1` and `du=dx`.

`int(x+2)/(x+1)dx=int((u-1)+2)/udu=int(u+1)/udu`

`=int1+1/udu=u+lnu+C=x+1+lnabs(x+1)+C`

The constant `1` is absorbed into the `C`, so the antiderivative is

`x+lnabs(x+1)+C`

Using another more intuitive approach:

`int(x+2)/(x+1)dx=int(x+1+1)/(x+1)dx=int(x+1)/(x+1)+1/(x+1)dx`

`=int1+1/(x+1)dx`

Again, substitution can be applied or you could realize that the derivative of `x+1` is `1`, so `int1/(x+1)dx` is `lnabs(x+1)` and `intdx=x`.

Instead of using the method of splitting `x+2` into `x+1+1`, polynomial long division confirms that `(x+2)/(x+1)=1+1/(x+1)`. Polynomial long division is always a good approach when the degree of the numerator equals the degree of the denominator.