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anonymous
 4 years ago
evaluating indefinite integral
anonymous
 4 years ago
evaluating indefinite integral

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{x^4 + 1}{x^2 \sqrt{x^4  1}}dx\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0\[u=x^41\] \[(u+1)^{1/4}=x\] \[\frac 14(u+1)^{3/4}\text du=\text dx\] \[\int\limits \frac{u+2}{(u+1)^{2/4} \sqrt{u}}\frac 14(u+1)^{3/4}\text du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have already tried that substitution and stuck

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.0maybe there is a trig substitution

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits\frac{(x^{2})^{2}+1}{x^{2}\sqrt{(x^{2})^{2}1}} \] maybe a trig sub..u^2 = tan\theta or u^2 = sin\theta

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1i mean x^2 = tan\theta or x^2 = sin\theta

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is guessing game...maybe\[x^2=\sec u\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1yeah my bad..but i think x^2 = tan(u) would work as well..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0here it is\[\int\limits \frac{\sec^2 u + 1}{2 \sqrt{\sec u}} du\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I don't know what to do with the sqrt{sec u}

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1secx = 1/cosx maybe weirstrass substitution?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1wait..wont work http://www.wolframalpha.com/input/?i=integrate+sqrt%28secx%29

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0there must be a tricky substitution

cruffo
 4 years ago
Best ResponseYou've already chosen the best response.0Have you tried splitting the integral up like this: ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it won't work... http://www.wolframalpha.com/input/?i=integrate+%28x^2%29%2F%28sqrt%28x^4+1%29%29+dx

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1then 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits \frac{\sec^2 u+1}{2\sqrt{\sec u}} du\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1hmm..this is hard..would integration by parts work?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0haven't tried that, but I doubt it will work

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0\[sec=\sqrt{tan^2+1}\]\[sec^2=tan^2+1\] \[\int\limits \frac{\tan^2 u+2}{2tan^2u+2} du\] or am i missing something?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0my last hope is x=e^t

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.1x=e^t? i might sound stupid; but where did e^t come from?

amistre64
 4 years ago
Best ResponseYou've already chosen the best response.0i missed something :)\[x^{1/4}\ne x\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well maybe that will lead us to some sinh and cosh

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x=e^t dx=e^t dt\[\int \frac{e^{4t} + 1}{e^{2t} \sqrt{e^{4t}  1}}e^t\text{d}t=\int \frac{e^{2t} + e^{2t}}{ \sqrt{e^{2t}  e^{2t}}}\text{d}t=\int \frac{2\cosh t}{\sqrt{\sinh t}} \text{d}t\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OOoopS typo again\[\int \frac{2\cosh 2t}{\sqrt{2\sinh 2t}} \text{d}t\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now I can rest peacefully lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ty all for helping :)
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