How do you integrate $10/(5x^2-2x^3)$ using partial fractions?

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Answer 1 · 2017-02-02T06:39:31

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

$5x^2-2x^3=-x^2(2x-5)$

Let's perform the decompostion into partial fractions

$10/(5x^2-2x^3)=-10/((x^2(2x-5)))$

$=A/x^2+B/x+C/(2x-5)$

$=(A(2x-5)+Bx(2x-5)+Cx^2)/((x^2(2x-5)))$

The denpminators are the same, we can compare the numerators

$-10=A(2x-5)+Bx(2x-5)+Cx^2$

Let $x=0$ , $=>$ , $-10=-5A$ , $=>$ , $A=2$

Let $x=5/2$ , $=>$ , $-10=25/4C$ , $=>$ , $C=-8/5$

Coefficients of $x^2$

$0=2B+C$ , $=>$ , $B=-C/2=1/2*8/5=4/5$

Therefore,

$10/(5x^2-2x^3)=2/x^2+(4/5)/x+(-8/5)/(2x-5)$

So,

$int(10dx)/(5x^2-2x^3)=2intdx/x^2+4/5intdx/x-8/5intdx/(2x-5)$

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

How do you integrate 10/(5x^2-2x^3)10/(5x^2-2x^3) using partial fractions?