Question

Integrate: dx/(x^2-4)^2

Answer 1

\[\int\limits_{}^{} (dx)/ (x^{2} - 4)^{2}\]

Answer 2

anyone? integration?

Answer 3

http://www.wolframalpha.com/input/?i=Integrate%3A+dx%2F%28x%5E2-4%29%5E2

Answer 4

hmm.. i thnk it should be (x-2)^2 and (x+2)^2
then solve it by \[A \div(x-2) + B \div (x-2)^{2}\]

Answer 5

and so on...

Answer 6

can you use trigonometric substitution? @cherrilyn

Answer 7

yeah im just not sure which trig sub I should use

Answer 8

i mean are you allowed to use

Answer 9

I'm trying to use trigo sub. It turns out to be something... not adorable :(

Answer 10

is wolframalpha's answer correct? I want to know the steps to reach the answer

Answer 11

hmmm yeah not pretty...i'll go with @vedic 's solution

\(\LARGE \frac{A}{x+2} + \frac{B}{x-2} + \frac{C}{(x+2)^2} + \frac{D}{(x-2)^2}\)

though it's not looking pretty either :/

Answer 12

@cherrilyn you can click the 'show steps' button

Answer 13

though @psujono you have to admit..wolfram's answers are crazy manipulations :p

Answer 14

If you use trigo. sub.
Let x= 2secu
dx = 2 secu tanu du
\[\int \frac{dx}{(x^2-4)^2} = \int \frac{2 secu tanudu}{((2secu)^2-4)^2}=\frac{1}{8} \int \frac{secu tanudu}{(tan^2u)^2} \]\[=\frac{1}{8} \int \frac{secu du}{(tan^3u)} =\frac{1}{8} \int cscucot^2u du = -\frac{1}{8} \int cotu d(cscu) \]\[= -\frac{1}{8} [cotucscu-\int cscu d(cotu)] =-\frac{1}{8} [cotucscu-\int csc^3u du] \]
Then I can't continue :|

Answer 15

Sorry, last step is: \[=-\frac{1}{8} [cotucscu+\int csc^3u d(u)]\]