Question

\int\limits \frac{ x^2-1 }{ x^3\sqrt{2x^4-2x^2+1} }

Answer 1

\[\int\limits \frac{ x^2-1 }{ x^3\sqrt{2x^4-2x^2+1} }dx\]

Answer 2

just trying.....
whats the derivative of
\(\Large \dfrac{2x^4-2x^2+1}{x^4}\)

did you get it something like 1/x-1/x^3 ...

Answer 3

if yes, then you can put that as u and see what happens

Answer 4

i started off with a substitution like x^2=t..but then failed

Answer 5

derivative of the above is ..\[\frac{ 4(x^2-1)}{ x^5 }\]

Answer 6

\(\Large \dfrac{2x^4-2x^2+1}{x^4} =2 - \dfrac{2}{x^2} + x^{-4} \\ d= 4/x -4/x^3 = 4 (x^2-1)/x^3 \)

Answer 7

umm ..yeahh ..i think i messed that up with u/v rule ..!

Answer 8

so we have the form
\(\Large (1/4)\int \dfrac{d(f(x))}{\sqrt {f(x)}}\)
form

Answer 9

no we don't ...i guess my method/trial failed....

Answer 10

yeahh ..i have the answer ...do u need that ? would dat be any help ?

Answer 11

i have the answer too.

Answer 12

umm do u have this .?\[\frac{ \sqrt{2x^4-2x^2+1} }{ 2x^2 } + c\]

Answer 13

yup

Answer 14

how did u get that ?

Answer 15

@myininaya @RadEn @ganeshie8 @mathmale might wanna have a look...

Answer 16

@ganeshie8 ??

Answer 17

\(\large \int\frac{ x^2-1 }{ x^3\sqrt{2x^4-2x^2+1} }dx\)

divide top and bottom wid x^2

\(\large \int\frac{ 1-\frac{1}{x^2} }{ x^3\sqrt{2-\frac{2}{x^2}+\frac{1}{x^4}} }dx\)

sub u = 1/x^2
du = -2/x^3 dx

\(\large \int\frac{ 1-u }{\sqrt{2-2u+u^2} } (-\frac{du}{2})\)

Answer 18

somplete the square in the bottom, and sub \(u-1 = \tan\theta\)

Answer 19

*complete

Answer 20

aah! one more substitution of u = 1/x^2 did the trick!
good catch @ganeshie8

Answer 21

:)

Answer 22

humm reverse trig sub can be tricky, if we want the given answer in terms of x's

Answer 23

amazing ....:!!!! thank you all .!

Answer 24

np :)