How do you evaluate the integral $int 1/(x(x-1)^3)$ ?

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Answer 1 · 2018-03-14T21:24:01

$int 1/(x(x-1)^3) dx = -ln abs(x)+ln abs(x-1)+1/(x-1)-1/(2(x-1)^2)+C$

Write:

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

Multiplying both sides by $x(x-1)^3$ we get:

$1 = A(x-1)^3+Bx(x-1)^2+Cx(x-1)+Dx$

Putting $x=0$ we find $A=-1$

Looking at the coefficient of $x^3$ , we find $B=-A = 1$

Putting $x=1$ we find $D=1$

Looking at the coefficient of $x$ , we find:

$0 = 3A+B-C+D$

Hence $C=-1$

So

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

and:

$int 1/(x(x-1)^3) dx = int -1/x+1/(x-1)-1/(x-1)^2+1/(x-1)^3 dx$

$color(white)(int 1/(x(x-1)^3) dx) = -ln abs(x)+ln abs(x-1)+1/(x-1)-1/(2(x-1)^2)+C$

How do you evaluate the integral int 1/(x(x-1)^3)int 1/(x(x-1)^3)?