Question

Trying to figure out if I did this correctly. If someone could look at my work it would be appreciated! :)

Indefinite integral of dx/x^2 (sqrt(1+x^2)

(I'll put my work in the comments section)

Answer 1

\[\int\frac{dx}{x^2\sqrt{1+x^2}}~~?\]

Answer 2

If \[a^2+u^2\] then u=a tan theta and \[1 + \tan \theta^2=\sec \theta^2\]
a=1 and u=x
x=tan theta and dx=sec^2theta
theta=arctan x
so
\[\int\limits (\sec \theta^2 d \theta)/\tan \theta^2 \sqrt (1+\tan \theta^2) \]
and then I ended up with
\[\int\limits 1+\tan \theta^2/ \tan \theta^2 (1+\tan \theta)\]

I cancelled out some things (not sure if I was supposed to) and ended up with
\[\int\limits 1/\tan \theta\]

and that is equal to \[1/\sec \theta^2\]

Answer 3

yes, that's the correct equation, SithsAndGiggles

Answer 4

Your substitution is right, but I get lost with what you do after that.
\[\begin{align*}\int\frac{\sec^2\theta}{\tan^2\theta\sqrt{1+\tan^2\theta}}~d\theta&=\int\frac{1+\tan^2\theta}{\tan^2\theta\sqrt{1+\tan^2\theta}}~d\theta\\
&=\int\frac{\sqrt{1+\tan^2\theta}}{\tan^2\theta}~d\theta\\
&=\int\frac{\sqrt{\sec^2\theta}}{\tan^2\theta}~d\theta\\
&=\int\frac{\sec\theta}{\tan^2\theta}~d\theta\\
&=\int\frac{\cos\theta}{\sin^2\theta}~d\theta
\end{align*}\]

Answer 5

let x=1/t

Answer 6

so i just realized my integration is NOT correct...but is the work up to that point even close?

Answer 7

and get

\[ I = -\int \frac{t}{\sqrt{t^2 + 1}}dt\]

Answer 8

^^The alternative method works so much more smoothly :)

Answer 9

Thank you both!!!

Answer 10

np