# Find the indefinite integral: {eq}\displaystyle \int \frac{x^2 + x - 1}{x^2 + x} \ dx {/eq}

## Question:

Find the indefinite integral:

{eq}\displaystyle \int \frac{x^2 + x - 1}{x^2 + x} \ dx {/eq}

## Solving Integration Using Partial Fraction:

To solve this kind of problems we will use the partial fraction i.e. {eq}\left(\dfrac A{x+a}+\dfrac B{x+b}+\ldots\right) {/eq} form. This way we will be able to apply the standard integration formula \begin{align} \int \dfrac{dx}{x+a}=\ln (x+a) \end{align}

Don't forget the integration constant.

We have to evaluate: {eq}\int \dfrac{x^2 + x - 1}{x^2 + x} \ dx {/eq}

\begin{align} \int \frac{x^2 + x - 1}{x^2 + x} \ dx&=\int \frac{(x^2 + x) - 1}{x^2 + x} \ dx\\[.3 cm] &=\int \left(\frac{x^2 + x }{x^2 + x}-\dfrac1{x^2+x} \right)\ dx\\[.3 cm] &=\int \left(1-\dfrac1{x(x+1)} \right)\ dx\\[.3 cm] &=\int \left[1-\left(\dfrac1x-\dfrac1{x+1} \right)\right]\ dx\\[.3 cm] &=\int \left(1-\dfrac1x+\dfrac1{x+1} \right)\ dx\\[.3 cm] &=x-\ln x+\ln (x+1)+C&\text{[C is an arbitrary integration constant.]} \end{align}

Hence, {eq}\begin{align} \boxed{\int \frac{x^2 + x - 1}{x^2 + x} \ dx=x-\ln x+\ln (x+1)+C} \end{align} {/eq}