How do you integrate $(x+7)/(x^2(x+2))$ using partial fractions?

Question

2717 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-08T18:04:26

The answer is $=-7/(2x)-5/4ln(|x|)+5/4ln(|x+2|)+C$

Let's perform the decomposition into partial fractions

$(x+7)/(x^2(x+2))=A/(x^2)+B/(x)+C/(x+2)$

$=(A(x+2)+B(x(x+2))+C(x^2))/(x^2(x+2))$

As the denominators are the same, we compare the numerators

$x+7=A(x+2)+B(x(x+2))+C(x^2)$

Let $A=0$ , $=>$ , $7=2A$ , $=>$ , $A=7/2$

Let $x=-2$ , $=>$ , $5=4C$ , $=>$ , $C=5/4$

Coefficients of $x$

$1=A+2B$ , $=>$ , $2B=1-A=1-7/2=-5/2$

$B=-5/4$

Therefore,

$(x+7)/(x^2(x+2))=(7/2)/(x^2)+(-5/4)/(x)+(5/4)/(x+2)$

So,

$int((x+7)dx)/(x^2(x+2))=7/2intdx/(x^2)-5/4intdx/(x)+5/4intdx/(x+2)$

$=-7/(2x)-5/4ln(|x|)+5/4ln(|x+2|)+C$

How do you integrate (x+7)/(x^2(x+2))(x+7)/(x^2(x+2)) using partial fractions?