Question

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

Answer 1

I think it should be \[\int\limits_{}^{} 1/\sin(x)\cos^2x\]

Answer 2

Or start from \[\int\frac{\sqrt{1+x^{2}}}{x}
dx\]

Answer 3

You could started from there, I got up to 1/sinxcos^2x

Answer 4

Assuming that you did the other part correct then..
\[\int \frac{1}{sinxcos^2{x}}\]
wait..are you sure about that though?

Answer 5

\[\sec \theta = \sqrt{1+ x^2}\]
\[\tan \theta = x\]
\[\sec^2 \theta d\theta = dx\]

\[\int \frac{\sqrt{1+x^2}}{x}dx \implies \frac{\sec \theta(\sec^2 \theta d\theta )}{\tan \theta}\]

Answer 6

I'm sure that \[\int\limits_{}^{} \sec^3x/\tan(x)\]

Answer 7

That turns into 1/cos^2xsinx

Answer 8

it does?

Answer 9

i mean yeah it does

Answer 10

I know I can't do u sub.

Answer 11

there's only one thing you can do...integration by parts

Answer 12

\[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \csc x \sec^2 x dx\]

Answer 13

But can I integrate that?

Answer 14

well i was just spitting out ideas....

Answer 15

That's why I didn't change it to that because I knew there was nothing to do there.

Answer 16

well i know i can make 1 = sin^2 x + cos^2 x

Answer 17

\[\int \frac{1}{\sin x \cos^2 x}dx \implies \int \frac{ \sin^2 x + \cos^2 x}{\sin x \cos^2 x}dx\]

Answer 18

OH! That's good

Answer 19

\[\int \frac{\sin^2 x}{\sin x \cos ^2 x}dx + \int \frac{\cos^2 x}{\sin x \cos^2 x}dx\]

Answer 20

yup saw the answer now :D i'll let you do the good stuff ;)

Answer 21

Ah, yes, yes, I see it too.

Answer 22

nice! congrats