How do you integrate $x^4/(x-1)^3$ using partial fractions?

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Answer 1 · 2017-12-30T07:26:04

The answer is $=-1/(2(x-1)^2)-4/(x-1)+6ln(|x-1|)+x^2/2+3x+C$

The degree of the numerator is greater than the degree of the denominator

Perform a polynomial long division

$x^4/(x-1)^3=(6x^2-8x+3)/(x-1)^3+x+3$

Now, we can perform the decomposition into partial fractions

$(6x^2-8x+3)/(x-1)^3=A/(x-1)^3+B/(x-1)^2+C/(x-1)$

$=(A+B(x-1)+C(x-1)^2)/(x-1)^3$

The denominators are the same, compare the numerators

$(6x^2-8x+3)=A+B(x-1)+C(x-1)^2$

Let $x=1$ , $=>$ , $A=1$

Let $x=0$ , $=>$ , $3=A-B+C$

$=>$ , $C-B=2$

Coefficients of $x^2$

$6=C$

$B=6-2=4$

Therefore,

$(6x^2-8x+3)/(x-1)^3=1/(x-1)^3+4/(x-1)^2+6/(x-1)$

So,

$x^4/(x-1)^3=1/(x-1)^3+4/(x-1)^2+6/(x-1)+x+3$

Finally,

$int(x^4dx)/(x-1)^3=int(dx)/(x-1)^3+4int(dx)/(x-1)^2+6int(dx)/(x-1)+intxdx+int3dx$

$=-1/(2(x-1)^2)-4/(x-1)+6ln(|x-1|)+x^2/2+3x+C$

How do you integrate x^4/(x-1)^3 x^4/(x-1)^3 using partial fractions?