# Use partial fractions to find these integrals: a. \int \frac{2x + 4}{x^3 - 2x^2} dx b. \int...

## Question:

Use partial fractions to find these integrals:

a. {eq}\int \frac{2x + 4}{x^3 - 2x^2} dx {/eq}

b. {eq}\int \frac{2x^2 - 5x + 2}{x^3 + x} dx {/eq}

## The partial fraction:

The partial fraction integration method is a form of integration that allows you to solve integrals of some kind of rational functions (ratio of polynomials), which could hardly be solved with other methods. Integration by partial fractions is an algebraic resource that will allow the resolution of this type of integrals.

## Answer and Explanation:

**Part a. **

{eq}\int \frac{2x + 4}{x^3 - 2x^2} dx {/eq}

** Resolve into partial fractions.**

{eq}\displaystyle \frac{2x + 4}{x^3 - 2x^2} \\ \displaystyle \frac{2x + 4}{x^2(x- 2)} = \frac {A}{x} +\frac {B}{x^2}+\frac {C}{x-2}\\ \displaystyle \frac{2x + 4}{x^2(x- 2)}= \frac {Ax(x-2)+B(x-2)+Cx^2}{x^2(x-2)} \\ \displaystyle \frac{2x + 4}{x^2(x- 2)}= \frac {Ax^2-2Ax+Bx-2B+Cx^2}{x^2(x-2)} \\ \displaystyle \frac{2x + 4}{x^2(x- 2)}= \frac {x^2(A+C)+x(-2A-2B)-2B}{ x^2(x-2) } \\ \displaystyle -2B=4 \Longrightarrow B=-2\\ \displaystyle -2A-2B=2 \Longrightarrow A=1\\ \displaystyle A+C =0 \Longrightarrow C=-1\\ \displaystyle \frac{2x + 4}{x^2(x- 2)} = \frac {1}{x} -\frac {2}{x^2}-\frac {1}{x-2} {/eq}

**Calculate the integral**

{eq}\displaystyle I = \int \frac{2x + 4}{x^3 - 2x^2} \, dx\\ \displaystyle I = \int \frac {1}{x} -\frac {2}{x^2}-\frac {1}{x-2} \, dx\\ \displaystyle I = \ln |x| +\frac {4}{x^3}- \ln |x-2| +C {/eq}

**The result of the integral applying partial fraction method is:**

{eq}\displaystyle \Longrightarrow \boxed{ I = \ln |x| +\frac {4}{x^3}- \ln |x-2| +C}\\ {/eq}

**Part b. **

{eq}\int \frac{2x^2 - 5x + 2}{x^3 + x} dx {/eq}

** Resolve into partial fractions.**

{eq}\displaystyle \frac{2x^2 - 5x + 2}{x^3 + x}\\ \displaystyle \frac{2x^2 - 5x + 2}{x(x^2+1)} = \frac {A}{x} +\frac {Bx+C}{x^2+1}\\ \displaystyle \frac{2x^2 - 5x + 2}{x(x^2+1)} = \frac {A(x^2+1)+(Bx+C)x }{x(x^2+1)} \\ \displaystyle \frac{2x^2 - 5x + 2}{x(x^2+1)} = \frac {Ax^2+A+Bx^2+Cx}{ x(x^2+1)} \\ \displaystyle \frac{2x^2 - 5x + 2}{x(x^2+1)} = \frac {x^2(A+B)+Cx+A}{(x+1)^3} \\ \displaystyle A=2\\ \displaystyle C=-5\\ \displaystyle A+B =2 \Longrightarrow B=0\\ \displaystyle \frac{2x^2 - 5x + 2}{x(x^2+1)} = \frac {2}{x} -\frac {5}{x^2+1} {/eq}

**Calculate the integral**

{eq}\displaystyle I = \int \frac{2x^2 - 5x + 2}{x^3 + x} \, dx\\ \displaystyle I = \int \frac {2}{x} -\frac {5}{x^2+1} \, dx\\ \displaystyle I = 2\ln |x| -5 \arctan (x)+C {/eq}

**The result of the integral applying partial fraction method is:**

{eq}\displaystyle \Longrightarrow \boxed{ I = 2\ln |x| -5\arctan (x)+C }\\ {/eq}