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#