4

$$ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $$ I don't know about decomposition of fractions a lot,but I know that it is method which I have to use in this example. Please, help me.

K.Hurwitz
  • 149
  • 2
    Is this a problem from a class? If so, then presumably partial fractions was discussed in class, or your book should have an explanation. Have you tried anything? – rogerl Dec 31 '15 at 14:28

2 Answers2

22

You have: $$ \int {\frac{(x-1)dx}{(x-2)(x+1)^2 x^2 }} $$ So it’s: $$ \frac{x-1}{(x-2)(x+1)^2 x^2} = \frac{a}{(x+1)^2} + \frac{b}{x+1} + \frac{c}{x^2} + \frac{d}{x} + \frac{e}{x-2} $$ $$ x-1 = ax^3-2ax^2+bx^2 (x^2+x-2x-2)+c(x-2)(x+1)^2+dx(x-2)(x+1)^2+e(x+1)^2x^2 $$ $$ x-1 = ax^3-2ax^2+bx^2(x^2+x-2x-2)+(cx-2c)(x^2+2x+1)+(dx^2-2dx)(x^2+2x+1)+ex^2(x^2+2x+1) $$ $$ x-1 = ax^3-2ax^2+bx^4-bx^3-2bx^2+cx^3+2cx^2+cx-2cx^2-4cx-2c+dx^4+2dx^3+dx^2-2dx^3-4dx^2-2dx+ex^4+2ex^3+ex^2 $$ $$ x-1 = ax^3-2ax^2+bx^4-bx^3-2bx^2+cx^3-3cx-2c+dx^4-3dx^2-2dx+ex^4+2ex^3+ex^2 $$ you have 5 equations: $$ \begin{cases} 0=b+d+e \\ 0=a-b+c+2e \\ 0=-2a-2b-3d+e \\ 1=-3c-2d \\ -1=-2c \end{cases} $$ from 5th equotion: $$ c= \frac{1}{2} $$ from 4th: $$ 1+2d=-3c,\quad2d=-3c-1,\quad 2d= \frac{-5}{2},\quad d= \frac{-5}{4} $$ from 1st: $$ b=-d-e $$ from 2nd: $$ a-(-d-e)+c+2e=0,\quad a+d+e+c+2e=0,\quad a+d+3e+c=0 $$ $$ a- \frac{5}{4} +3e + \frac{1}{2} = 0, \quad a+3e= \frac{3}{4}, \quad a=\frac{3}{4} -3e $$ from 3rd: $$ 0=-2(\frac{3}{4}-3e) -2(-d-e) -3d+e,\quad 0=-\frac{6}{4} + 6e+2d+2e-3d+e $$ $$ 0=-\frac{6}{4} + 6e - \frac{10}{4} +2e+ \frac{15}{4} +e,\quad 0=-\frac{1}{4} +9e $$ $$ e=\frac{1}{36}, \quad a=\frac{3}{4}-3e=\frac{3}{4}-\frac{3}{36}=\frac{24}{35} , \quad a=\frac{2}{3} $$ $$ b=-d-e=\frac{5}{4}-\frac{1}{36}=\frac{44}{36},\quad b=\frac{11}{9} $$ so you have: $$ \begin{cases} a=\frac{2}{3} \\ b=\frac{11}{9} \\ c= \frac{1}{2} \\ d= \frac{-5}{4} \\ e=\frac{1}{36} \end{cases} $$ and now: $$ \int \left( \frac{2}{3}\frac{1}{(x+1)^2} + \frac{11}{9}\frac{1}{x+1} + \frac{1}{2}\frac{1}{x^2} - \frac{5}{4}\frac{1}{x} + \frac{1}{36}\frac{1}{x-2} \right)dx = -\frac{2}{3}\frac{1}{x+1} + \frac{11}{9}\ln|x+1| - \frac{1}{2x}-\frac{5}{4}\ln|x| + \frac{1}{36}\ln|x-2| +C $$ so the answer is: $$ \color{red}{-\frac{2}{3}\frac{1}{x+1}+\frac{11}{9}\ln|x+1|-\frac{1}{2x}-\frac{5}{4}\ln|x|+\frac{1}{36}\ln|x-2|+C} $$

Kamil Jarosz
  • 4,984
A.Łabuś
  • 244
0

$$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2} $$ $$ \frac{x-1}{ x^2(x+1)^2} = A + \frac{Bx+C}{x^2}(x-2) + \frac{D(x+1)+E}{(x+1)^2}(x-2). $$ $x\to 2: A = \frac1{36}.$ $$ \frac{x-1}{(x-2)(x+1)^2} = \frac A{x-2}x^2+Bx+C + \frac{D(x+1)+E}{(x+1)^2}x^2 $$ $x\to 0: C = \frac12.$ $$ \frac{x-1}{(x-2)x^2} = \frac A{x-2}(x+1)^2+\frac{Bx+C}{x^2}(x+1)^2 + D(x+1)+E$$ $x\to -1: E = \frac23.$ $$ \frac{x-1}{(x-2)x(x+1)^2} = \frac A{x-2}x+\frac{Bx+C}{x} + \frac{D(x+1)+E}{(x+1)^2}x $$ $x\to\infty: A + B + D = 0.\quad (1)$ $$ \frac{x-1}{(x-2) x^2 (x+1)^2} = \frac A{x-2}+\frac{Bx+C}{x^2} + \frac{D(x+1)+E}{(x+1)^2}$$ $x\to 1: -A+B+C+\frac{2D+E}4 = 0. \quad (2)$ $$(1)-(2): 2A-C+ \frac D2-\frac E4 = 0,$$ $$D=-4A+2C+\dfrac E2 = -\frac19+1+\frac13 = \frac{11}9$$ $B=-(D+A) = -(\frac{11}9+\frac1{36}) = -\frac 54$

Yuri Negometyanov
  • 28,026