Question

evaluating indefinite integral

Answer 1

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

Answer 2

\[u=x^4-1\]
\[(u+1)^{1/4}=x\]
\[\frac 14(u+1)^{-3/4}\text du=\text dx\]

\[\int\limits \frac{u+2}{(u+1)^{2/4} \sqrt{u}}\frac 14(u+1)^{-3/4}\text du\]

Answer 3

hmm

Answer 4

I have already tried that substitution and stuck

Answer 5

maybe there is a trig substitution

Answer 6

\[\int\limits\frac{(x^{2})^{2}+1}{x^{2}\sqrt{(x^{2})^{2}-1}} \]
maybe a trig sub..u^2 = tan\theta or u^2 = sin\theta

Answer 7

i mean x^2 = tan\theta or x^2 = sin\theta

Answer 8

this is guessing game...maybe\[x^2=\sec u\]

Answer 9

yeah my bad..but i think x^2 = tan(u) would work as well..

Answer 10

here it is\[\int\limits \frac{\sec^2 u + 1}{2 \sqrt{\sec u}} du\]

Answer 11

I don't know what to do with the sqrt{sec u}

Answer 12

secx = 1/cosx

maybe weirstrass substitution?

Answer 13

wait..wont work
http://www.wolframalpha.com/input/?i=integrate+sqrt%28secx%29

Answer 14

there must be a tricky substitution

Answer 15

Have you tried splitting the integral up like this:
?

Answer 16

it won't work... http://www.wolframalpha.com/input/?i=integrate+%28x^2%29%2F%28sqrt%28x^4+-1%29%29+dx

Answer 17

then what about this \[\int\limits\frac{(secu)^{2}+1}{(secu)^{2}\sqrt{(secu)^{2}-1}} *\frac{tanusecu}{2\sqrt{secu}} \]?
its not going to look nice tho

Answer 18

\[\int\limits \frac{\sec^2 u+1}{2\sqrt{\sec u}} du\]

Answer 19

hmm..this is hard..would integration by parts work?

Answer 20

haven't tried that, but I doubt it will work

Answer 21

\[sec=\sqrt{tan^2+1}\]\[sec^2=tan^2+1\]
\[\int\limits \frac{\tan^2 u+2}{2tan^2u+2} du\]
or am i missing something?

Answer 22

my last hope is x=e^t

Answer 23

x=e^t? i might sound stupid; but where did e^t come from?

Answer 24

i missed something :)\[x^{1/4}\ne x\]

Answer 25

well maybe that will lead us to some sinh and cosh

Answer 26

yeah x=e^t works

Answer 27

x=e^t

dx=e^t dt\[\int \frac{e^{4t} + 1}{e^{2t} \sqrt{e^{4t} - 1}}e^t\text{d}t=\int \frac{e^{2t} + e^{-2t}}{ \sqrt{e^{2t} - e^{-2t}}}\text{d}t=\int \frac{2\cosh t}{\sqrt{\sinh t}} \text{d}t\]

Answer 28

OOoopS typo again\[\int \frac{2\cosh 2t}{\sqrt{2\sinh 2t}} \text{d}t\]

Answer 29

nice, it works!

Answer 30

:)

Answer 31

now I can rest peacefully lol

Answer 32

ty all for helping :)