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

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Answer 1 · 2016-11-19T07:12:51

The answer is $= \frac{2}{5 x} - \frac{7}{5 (x + 5)}$ $=2/(5x)-7/(5(x+5))$

Let's go to the decomposition in partial fractions

$\frac{2 - x}{x^{2} + 5 x} = \frac{2 - x}{(x) (x + 5)}$ $(2-x)/(x^2+5x)=(2-x)/((x)(x+5))$

$= \frac{A}{x} + \frac{B}{x + 5} = \frac{A (x + 5) + B x}{x (x + 5)}$ $=A/x+B/(x+5)=(A(x+5)+Bx)/(x(x+5))$

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

Let $x = 0$ $x=0$ , $\Rightarrow$ $=>$ $2 = 5 A$ $2=5A$ ; $\Rightarrow$ $=>$ , $A = \frac{2}{5}$ $A=2/5$

Let $x = - 5$ $x=-5$ ; $\Rightarrow$ $=>$ , $7 = - 5 B$ $7=-5B$ ; $\Rightarrow$ $=>$ , $B = - \frac{7}{5}$ $B=-7/5$

$\frac{2 - x}{x^{2} + 5 x} = \frac{\frac{2}{5}}{x} + \frac{- \frac{7}{5}}{x + 5} =$ $(2-x)/(x^2+5x)=(2/5)/x+(-7/5)/(x+5)=$

How do you integrate 2−xx2+5x(2-x)/(x^2+5x) using partial fractions?