How do I find the integral $\int \frac{d x}{x^{2} {(x - 1)}^{2}}$ $intdx/(x^2(x-1)^2)$ ?

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How do I find the integral $\int \frac{d x}{x^{2} {(x - 1)}^{2}}$ $intdx/(x^2(x-1)^2)$ ?

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Answer 1 · 2014-09-28T11:59:18

By Partial Fraction Decomposition,

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

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

by the partial fractions above,

$= \int \frac{2}{x} + \frac{1}{x^{2}} + \frac{2}{x - 1} + \frac{1}{{(x - 1)}^{2}} d x$ $=int2/x+1/x^2+2/{x-1}+1/{(x-1)^2}dx$

by Log Rule and Power Rule,

$= 2 ln | x | - \frac{1}{x} + 2 ln | x - 1 | - \frac{1}{x - 1} + C$ $=2ln|x|-1/x+2ln|x-1|-1/{x-1}+C$

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How do I find the integral $\int \frac{d x}{x^{2} {(x - 1)}^{2}}$ $intdx/(x^2(x-1)^2)$ ?

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