# How do you integrate int (x-1)/( x^4 (x-1)^2) using partial fractions?

Sep 9, 2017

$L n \left(x - 1\right) - L n x + {x}^{- 1} + \frac{1}{2} \cdot {x}^{- 2} + \frac{1}{3} \cdot {x}^{- 3} + C$

#### Explanation:

1) I used basic fractions method for integration.

$\int \frac{\left(x - 1\right) \mathrm{dx}}{{x}^{4} \cdot {\left(x - 1\right)}^{2}}$

$= \int \frac{\mathrm{dx}}{{x}^{4} \cdot \left(x - 1\right)}$

I decomposed integrand into basic fractions,

$\frac{1}{{x}^{4} \cdot \left(x - 1\right)} = \frac{A}{x - 1} + \frac{B}{x} + \frac{C}{x} ^ 2 + \frac{D}{x} ^ 3 + \frac{E}{x} ^ 4$

$A \cdot {x}^{4} + B \cdot {x}^{3} \cdot \left(x - 1\right) + C \cdot {x}^{2} \cdot \left(x - 1\right) + D \cdot x \cdot \left(x - 1\right) + E \cdot \left(x - 1\right) = 1$

Set $x = 0$, $- E = - 1$ or $E = - 1$

Set $x = 1$, $A = 1$

I took differentiaton both sides,

$4 A \cdot {x}^{3} + B \cdot \left(4 {x}^{3} - 3 {x}^{2}\right) + C \cdot \left(3 {x}^{2} - 2 x\right) + D \cdot \left(2 x - 1\right) + E = 0$

Set $x = 0$, $- D + E = 0$ or $D = E = - 1$

I took differentiaton both sides,

$12 A \cdot {x}^{2} + B \cdot \left(12 {x}^{2} - 6 x\right) + C \cdot \left(6 x - 2\right) + 2 D = 0$

Set $x = 0$, $- 2 C + 2 D = 0$ or $C = D = - 1$

I took differentiaton both sides,

$24 A \cdot x + B \cdot \left(24 x - 6\right) + 6 C = 0$

Set $x = 0$, $- 6 B + 6 C = 0$ or $B = C = - 1$

Thus,

$\int \frac{\mathrm{dx}}{{x}^{4} \cdot \left(x - 1\right)}$

=$\int \frac{\mathrm{dx}}{x - 1}$-$\int \frac{\mathrm{dx}}{x}$-$\int \frac{\mathrm{dx}}{x} ^ 2$-$\int \frac{\mathrm{dx}}{x} ^ 3$-$\int \frac{\mathrm{dx}}{x} ^ 4$

=$L n \left(x - 1\right) - L n x + {x}^{- 1} + \frac{1}{2} \cdot {x}^{- 2} + \frac{1}{3} \cdot {x}^{- 3} + C$

Sep 9, 2017

How about NOT using partial fractions and instead using the binomial theorem (or the sum of geometrical series, same thing) after cancelling the $x - 1$?

#### Explanation:

$\int \frac{1}{{x}^{4} \left(x - 1\right)} \mathrm{dx}$
$= - \int \frac{1}{{x}^{4} \left(1 - x\right)} \mathrm{dx}$
$= - \int \left(\frac{1}{x} ^ 4\right) \left(1 + x + {x}^{2} + {x}^{3} + {x}^{4.} . .\right) \mathrm{dx}$ (for suitable $x$, sum to infinity of GP with common ratio $x$, backwards, or binomial expansion with $n = - 1$)
$= - \int {x}^{-} 4 + {x}^{-} 3 + {x}^{-} 2 + {x}^{-} 1 + 1 + x + {x}^{2} + {x}^{3} + \ldots \mathrm{dx}$
$= \frac{1}{3} {x}^{3} + \frac{1}{2} {x}^{-} 2 + {x}^{-} 1 - \ln | x | - \int \frac{1}{1 - x} \mathrm{dx}$ (sum to infinity of GP again, forwards)
$= \frac{1}{3} {x}^{3} + \frac{1}{2} {x}^{-} 2 + {x}^{-} 1 - \ln | x | + \ln | x - 1 | + c$

The binomial expansion above works only for $| x | < 1$. To deal with $| x | > 1$:
$\int \frac{1}{{x}^{4} \left(x - 1\right)} \mathrm{dx}$
$= \int \frac{1}{{x}^{5} \left(1 - \frac{1}{x}\right)} \mathrm{dx}$
Using the Binomial Theorem, or sum to infinity of GP with first term 1 and common ratio ${x}^{-} 1$:
$= \int {x}^{-} 5 \left(1 + {x}^{-} 1 + {x}^{-} 2 + {x}^{-} 3. . .\right) \mathrm{dx}$ provided $| x | > 1$
$= \int {x}^{-} 5 + {x}^{-} 6 + {x}^{-} 7. . . \mathrm{dx}$
$= \int {x}^{-} 1 + {x}^{-} 2 + {x}^{-} 3 + {x}^{-} 4 + {x}^{-} 5. . . \mathrm{dx} -$
$\setminus \setminus \setminus \setminus \setminus \int {x}^{-} 1 + {x}^{-} 2 + {x}^{-} 3 + {x}^{-} 4 \mathrm{dx}$

Using sum to infinity of GP or binomial theorem:
$= \int {x}^{-} \frac{1}{1 - {x}^{-} 1} \mathrm{dx} - \int {x}^{-} 1 + {x}^{-} 2 + {x}^{-} 3 + {x}^{-} 4 \mathrm{dx}$
$= \int \frac{1}{x - 1} \mathrm{dx} - \int {x}^{-} 1 + {x}^{-} 2 + {x}^{-} 3 + {x}^{-} 4 \mathrm{dx}$
$= \ln | x - 1 | - \ln | x | + \frac{1}{x} + \frac{1}{2} {x}^{2} + \frac{1}{3} {x}^{3} + c$
which is the same as the result for $| x | < 1$. Good!