5x^2-2x^3=-x^2(2x-5)5x2−2x3=−x2(2x−5)
Let's perform the decompostion into partial fractions
10/(5x^2-2x^3)=-10/((x^2(2x-5)))105x2−2x3=−10(x2(2x−5))
=A/x^2+B/x+C/(2x-5)=Ax2+Bx+C2x−5
=(A(2x-5)+Bx(2x-5)+Cx^2)/((x^2(2x-5)))=A(2x−5)+Bx(2x−5)+Cx2(x2(2x−5))
The denpminators are the same, we can compare the numerators
-10=A(2x-5)+Bx(2x-5)+Cx^2−10=A(2x−5)+Bx(2x−5)+Cx2
Let x=0x=0, =>⇒, -10=-5A−10=−5A, =>⇒, A=2A=2
Let x=5/2x=52, =>⇒, -10=25/4C−10=254C, =>⇒, C=-8/5C=−85
Coefficients of x^2x2
0=2B+C0=2B+C, =>⇒, B=-C/2=1/2*8/5=4/5B=−C2=12⋅85=45
Therefore,
10/(5x^2-2x^3)=2/x^2+(4/5)/x+(-8/5)/(2x-5)105x2−2x3=2x2+45x+−852x−5
So,
int(10dx)/(5x^2-2x^3)=2intdx/x^2+4/5intdx/x-8/5intdx/(2x-5)∫10dx5x2−2x3=2∫dxx2+45∫dxx−85∫dx2x−5
=-2/x+4/5ln(|x|)-4/5ln(|2x-5|)+C=−2x+45ln(|x|)−45ln(|2x−5|)+C
