What is the integrate of x^2/(x-1)^3 ?

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What is the integrate of x^2/(x-1)^3 ?

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Answer 1 · 2018-04-09T09:02:31

$\int \frac{x^{2}}{{(x - 1)}^{3}} d x = log | x - 1 | + \frac{2}{1 - x} - \frac{1}{2 {(x - 1)}^{2}} + C$ $int x^2/(x-1)^3 dx=log|x-1|+2/(1-x)-1/(2(x-1)^2)+C$

Explanation:

Part 1

Decompose $\frac{x^{2}}{{(x - 1)}^{3}}$ $x^2/(x-1)^3$ into partial fractions

$\frac{x^{2}}{{(x - 1)}^{3}} = \frac{A}{x - 1} + \frac{B}{{(x - 1)}^{2}} + \frac{C}{{(x - 1)}^{3}}$ $x^2/(x-1)^3=A/(x-1)+B/(x-1)^2+C/(x-1)^3$

Where A, B and C are just numbers

So to solve, just merge those fractions

$\frac{x^{2}}{{(x - 1)}^{3}} = \frac{A {(x - 1)}^{2} + B (x - 1) + C}{{(x - 1)}^{3}}$ $x^2/(x-1)^3=(A(x-1)^2+B(x-1)+C)/(x-1)^3$
$= \frac{A x^{2} - 2 A x + A + B x - B + C}{{(x - 1)}^{3}}$ $=(Ax^2-2Ax+A+Bx-B+C)/(x-1)^3$

Rearranging

$\frac{x^{2}}{{(x - 1)}^{3}} = \frac{A x^{2} + (B - 2 A) x + (A - B + C)}{{(x - 1)}^{3}}$ $x^2/(x-1)^3=(Ax^2+(B-2A)x+(A-B+C))/(x-1)^3$

By comparing coefficients, you can tell that

$A = 1, B - 2 A = 0, A - B + C = 0$ $A=1, B-2A=0, A-B+C=0$

Solving, you get $A = 1, B = 2 and C = 1$ $A=1, B=2 and C=1$

Hence $\frac{x^{2}}{{(x - 1)}^{3}} = \frac{1}{x - 1} + \frac{2}{{(x - 1)}^{2}} + \frac{1}{{(x - 1)}^{3}}$ $x^2/(x-1)^3=1/(x-1)+2/(x-1)^2+1/(x-1)^3$

Part 2

Substituting this back into the integral, you get

$\int \frac{x^{2}}{{(x - 1)}^{3}} d x = \int \frac{1}{x - 1} d x + \int \frac{2}{{(x - 1)}^{2}} d x + \int \frac{1}{{(x - 1)}^{3}} d x$ $int x^2/(x-1)^3 dx=int 1/(x-1) dx+int 2/(x-1)^2 dx+int 1/(x-1)^3 dx$

$\int \frac{1}{x - 1} d x = log | x - 1 | + C$ $int 1/(x-1) dx=log|x-1|+C$

Using the reverse power rule

$\int \frac{1}{{(x - 1)}^{2}} d x = \int {(x - 1)}^{- 2} d x = - \frac{1}{x - 1} + C = \frac{1}{1 - x} + C$ $int 1/(x-1)^2 dx=int (x-1)^-2 dx=-1/(x-1)+C=1/(1-x)+C$

$\int \frac{1}{{(x - 1)}^{3}} d x = \int {(x - 1)}^{- 3} d x = - \frac{1}{2 {(x - 1)}^{2}} + C$ $int 1/(x-1)^3 dx=int (x-1)^-3 dx=-1/(2(x-1)^2)+C$

Hence
$\int \frac{x^{2}}{{(x - 1)}^{3}} d x = log | x - 1 | + \frac{2}{1 - x} - \frac{1}{2 {(x - 1)}^{2}} + C$ $int x^2/(x-1)^3 dx=log|x-1|+2/(1-x)-1/(2(x-1)^2)+C$

You can further simplify this to be

$\int \frac{x^{2}}{{(x - 1)}^{3}} d x = log | x - 1 | + \frac{2}{1 - x} - \frac{1}{2 {(x - 1)}^{2}} + C = \frac{3 - 4 x}{2 {(x - 1)}^{2}} + log | x - 1 | + C$ $int x^2/(x-1)^3 dx=log|x-1|+2/(1-x)-1/(2(x-1)^2)+C=(3-4x)/(2(x-1)^2)+log|x-1|+C$

P.S. Please classify this question as "Calculus", you are currently marking it as "algebra"

Answer 2 · 2018-04-09T09:19:06

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

Explanation:

Have a look at this Second Solution :

Let, $I = \int \frac{x^{2}}{{(x - 1)}^{3}} d x$ $I=intx^2/(x-1)^3dx$ .

We subst. $x=1+t. :. dx=dt, and, (x-1)=t$ .

$:. I=int(t+1)^2/t^3dt=int(t^2+2t+1)/t^3dt$ ,

$=int{t^2/t^3+2*t/t^3+1/t^3}dt$ ,

$=int{1/t+2*t^-2+t^-3}dt$ ,

$=ln|t|+2*t^(-2+1)/(-2+1)+t^(-3+1)/(-3+1)$ ,

$=ln|t|-2*t^-1-1/2*t^-2$ .

$rArr I=ln|(x-1)|-2/(x-1)-1/(2(x-1)^2)+C$ , as Respected

e52fa787 has already derived!

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What is the integrate of x^2/(x-1)^3 ?

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