How do you use partial fractions to find the integral $int (5x^2-12x-12)/(x^3-4x)dx$ ?

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Answer 1 · 2017-11-04T23:57:23

$int ((5x^2-12x-12)*dx)/(x^3-4x)$ = $4Ln(x+2)+3Lnx-2Ln(x-2)+C$

$(5x^2-12x-12)/(x^3-4x)$

= $(5x^2-12x-12)/[x*(x+2)*(x-2)]$

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

After expanding denominator,

$A*x*(x-2)+B*(x+2)*(x-2)+C*x*(x+2)=5x^2-12x-12$

Set $x=-2$ , $8A=32$ , so $A=4$

Set $x=0$ , $-4B=-12$ , so $B=3$

Set $x=2$ , $-8C=-16$ , so $C=-2$

Thus,

$int ((5x^2-12x-12)*dx)/(x^3-4x)$

= $int (4dx)/(x+2)$ + $int (3dx)/x$ - $int (2dx)/(x-2)$

= $4Ln(x+2)+3Lnx-2Ln(x-2)+C$

How do you use partial fractions to find the integral int (5x^2-12x-12)/(x^3-4x)dxint (5x^2-12x-12)/(x^3-4x)dx?