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.

Answer and Explanation: 1

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}


Learn more about this topic:

Loading...
How to Integrate Functions With Partial Fractions

from

Chapter 13 / Lesson 10
4.6K

Learn about integration by partial fractions. Explore how to make partial fractions and then how to integrate fractions. See examples of integrating fractions.


Related to this Question

Explore our homework questions and answers library