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Integrate. sqrt(1+x^2)/x

Mathematics
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I think it should be \[\int\limits_{}^{} 1/\sin(x)\cos^2x\]
Or start from \[\int\frac{\sqrt{1+x^{2}}}{x} dx\]
You could started from there, I got up to 1/sinxcos^2x

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Other answers:

Assuming that you did the other part correct then.. \[\int \frac{1}{sinxcos^2{x}}\] wait..are you sure about that though?
|dw:1342236036378:dw| \[\sec \theta = \sqrt{1+ x^2}\] \[\tan \theta = x\] \[\sec^2 \theta d\theta = dx\] \[\int \frac{\sqrt{1+x^2}}{x}dx \implies \frac{\sec \theta(\sec^2 \theta d\theta )}{\tan \theta}\]
I'm sure that \[\int\limits_{}^{} \sec^3x/\tan(x)\]
That turns into 1/cos^2xsinx
it does?
i mean yeah it does
I know I can't do u sub.
there's only one thing you can do...integration by parts
\[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \csc x \sec^2 x dx\]
But can I integrate that?
well i was just spitting out ideas....
That's why I didn't change it to that because I knew there was nothing to do there.
well i know i can make 1 = sin^2 x + cos^2 x
\[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \frac{ \sin^2 x + \cos^2 x}{\sin x \cos^2 x}dx\]
OH! That's good
\[\int \frac{\sin^2 x}{\sin x \cos ^2 x}dx + \int \frac{\cos^2 x}{\sin x \cos^2 x}dx\]
yup saw the answer now :D i'll let you do the good stuff ;)
Ah, yes, yes, I see it too.
nice! congrats

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