How do you integrate $(2-x)/(x^2+5x)$ using partial fractions?

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Answer 1 · 2016-11-19T07:12:51

The answer is $=2/(5x)-7/(5(x+5))$

Let's go to the decomposition in partial fractions

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

$=A/x+B/(x+5)=(A(x+5)+Bx)/(x(x+5))$

so, $(2-x)=A(x+5)+Bx$

Let $x=0$ , $=>$ $2=5A$ ; $=>$ , $A=2/5$

Let $x=-5$ ; $=>$ , $7=-5B$ ; $=>$ , $B=-7/5$

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

How do you integrate (2-x)/(x^2+5x)(2-x)/(x^2+5x) using partial fractions?