How do you integrate $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2 / (x-1)^3$ using partial fractions?

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How do you integrate $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2 / (x-1)^3$ using partial fractions?

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Answer 1 · 2016-12-07T17:57:10

$ln (x - 1) - \frac{2}{x - 1} - \frac{1}{2 {(x - 1)}^{2}} + C$ $ln(x-1)-2/(x-1)-1/(2(x-1)^2)+C$

Explanation:

Assume that there exist $A, B and C$ $A, B and C$ such that:
$\frac{x^{2}}{{(x - 1)}^{3}} \equiv \frac{A}{(x - 1)} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}}$ $x^2/(x-1)^3-=A/((x-1))+B/(x-1)^2+C/(x-1)^3$ .
To find $A$ $A$ , $B$ $B$ and $C$ $C$ substitute any three different numbers other than $- 1$ $-1$ and solve the resulting set of three linear equations. For example, set $x = 0$ $x=0$ , $2$ $2$ and $- 1$ $-1$ :
$0 = \frac{A}{- 1} + B + \frac{C}{- 1}$ $0=A/-1+B+C/-1$
$4 = A + B + C$ $4=A+B+C$
$- \frac{1}{8} = \frac{A}{- 2} + \frac{B}{4} - \frac{C}{8}$ $-1/8=A/-2+B/4-C/8$
giving $A = 1$ $A=1$ , $B = 2$ $B=2$ and $C = 1$ $C=1$ .
So the required integral is:
$\int \frac{1}{x - 1} + \frac{2}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}} d x$ $int1/(x-1)+2/(x-1)^2+1/(x-1)^3dx$
which gives the answer above.

Alternatively, instead of using partial fractions, substitute $u = x - 1$ $u=x-1$ to get $\int \frac{1}{u} + \frac{2}{u^{2}} + \frac{1}{u^{3}} d u$ $int1/u+2/u^2+1/u^3du$ .

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How do you integrate $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2 / (x-1)^3$ using partial fractions?

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How do you integrate $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2 / (x-1)^3$ using partial fractions?