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

Jan 28, 2016

$x + \ln \left\mid x + 1 \right\mid + C$

#### Explanation:

Using substitution:

Set $u = x + 1$. This means that $x = u - 1$ and $\mathrm{du} = \mathrm{dx}$.

$\int \frac{x + 2}{x + 1} \mathrm{dx} = \int \frac{\left(u - 1\right) + 2}{u} \mathrm{du} = \int \frac{u + 1}{u} \mathrm{du}$

$= \int 1 + \frac{1}{u} \mathrm{du} = u + \ln u + C = x + 1 + \ln \left\mid x + 1 \right\mid + C$

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

$x + \ln \left\mid x + 1 \right\mid + C$

Using another more intuitive approach:

$\int \frac{x + 2}{x + 1} \mathrm{dx} = \int \frac{x + 1 + 1}{x + 1} \mathrm{dx} = \int \frac{x + 1}{x + 1} + \frac{1}{x + 1} \mathrm{dx}$

$= \int 1 + \frac{1}{x + 1} \mathrm{dx}$

Again, substitution can be applied or you could realize that the derivative of $x + 1$ is $1$, so $\int \frac{1}{x + 1} \mathrm{dx}$ is $\ln \left\mid x + 1 \right\mid$ and $\int \mathrm{dx} = x$.

Instead of using the method of splitting $x + 2$ into $x + 1 + 1$, polynomial long division confirms that $\frac{x + 2}{x + 1} = 1 + \frac{1}{x + 1}$. Polynomial long division is always a good approach when the degree of the numerator equals the degree of the denominator.