## exraven 3 years ago evaluating indefinite integral

1. exraven

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

2. UnkleRhaukus

$u=x^4-1$ $(u+1)^{1/4}=x$ $\frac 14(u+1)^{-3/4}\text du=\text dx$ $\int\limits \frac{u+2}{(u+1)^{2/4} \sqrt{u}}\frac 14(u+1)^{-3/4}\text du$

3. UnkleRhaukus

hmm

4. exraven

I have already tried that substitution and stuck

5. UnkleRhaukus

maybe there is a trig substitution

6. Mimi_x3

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

7. Mimi_x3

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

8. mukushla

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

9. Mimi_x3

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

10. exraven

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

11. exraven

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

12. Mimi_x3

secx = 1/cosx maybe weirstrass substitution?

13. Mimi_x3
14. exraven

there must be a tricky substitution

15. cruffo

Have you tried splitting the integral up like this: ?

16. exraven
17. Mimi_x3

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

18. exraven

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

19. Mimi_x3

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

20. exraven

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

21. amistre64

$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?

22. mukushla

my last hope is x=e^t

23. Mimi_x3

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

24. amistre64

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

25. mukushla

well maybe that will lead us to some sinh and cosh

26. mukushla

yeah x=e^t works

27. mukushla

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

28. mukushla

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

29. exraven

nice, it works!

30. mukushla

:)

31. exraven

now I can rest peacefully lol

32. exraven

ty all for helping :)