Antiderivative for 1/[x²√(1+x²)]?

[Zorodius]

The title sums it up. How can I find this?

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

I've tried a lot of different things to come up with an answer. My answers must be wrong, but I don't see where I'm making a mistake. Here's one way I've tried to solve the problem:

Let
$u = (1+x^2)^{-1/2}$
$v = -x^{-1}$

Then
$u' = -x(1+x^2)^{-3/2}$
$v' = x^{-2}$

And so the original problem is equivalent to:

$$\int uv' = uv - \int u'v$$
$$= -\frac {1}{x (1+x^2)^{1/2}} - \int \frac{(-x)(-x^{-1}) dx}{(1+x^2)^{3/2}}$$
$$= -\frac {1}{x (1+x^2)^{1/2}} - \int \frac{dx}{(1+x^2)^{3/2}}$$

... which calls for this trigonometric substitution:

tan y = x
$$\frac{dy}{\cos^2 y} = dx$$

So, with that substitution, the original problem is equal to this:

$$-\frac {1}{x (1+x^2)^{1/2}} - \int \frac{dy}{\cos^2 y(\frac{1}{\cos^2 y})^{3/2}}$$
$$= -\frac {1}{x (1+x^2)^{1/2}} - \int \cos y \; dy$$

which is

$$= -\frac {1}{x (1+x^2)^{1/2}} - \sin \arctan x + C$$

The book's answer doesn't have any trigonometric functions in it, and more to the point, differentiating my answer doesn't give me the function inside the original problem, so that's out.

Where did I go wrong?

 

[whozum]

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

Try the trig substitution in the beginning.
$$x = tan u, dx = sec^2udu$$

$$\int\frac{sec^2u}{\tan^2usecu}du = \int \frac{secu}{\tan^2u}du = \int \frac{1}{\cos u \tan^2u}du = \int \frac{1}{\frac{sin^2u}{cos^2u}cosu}du = \int \frac{cos u}{sin^2u}du$$

A nice substitution will take care of that, and when your done with the simplifying you should have a trig function. Draw a triangle and figure out the final result which should be

$$\frac{-\sqrt{x^2+1}}{x}$$

 

[arildno]

Note that:
$$x=tan(arctan(x))=\frac{\sin(arctan(x))}{\sqrt{1-\sin^{2}(arctan(x))}}$$
That is, we get:
$$sin(arctan(x))=\frac{x}{\sqrt{1+x^{2}}}$$

Try and see if this is what you needed..

 

[arildno]

Note that your own answer agrees with whozum's by using the expression provided for sin(arctan(x))

 

[whozum]

Mine is simpler! :)

 

[arildno]

Sure enough.
Personally, however, I would have used the substitution x=Sinh(u) from the start, rather than a trigonometric substitution.

 

[whozum]

I don't know hyperbolic trigonometry, can you believe I am a physics junior and have never been taught it?

 

[Zorodius]

Thanks SO much! I thought there might be something I could substitute for sin atan x, but I didn't figure out what it should be.

 

[arildno]

I don't know hyperbolic trigonometry, can you believe I am a physics junior and have never been taught it?

You'll soon learn it; it isn't that difficult.

 

[arildno]

Thanks SO much! I thought there might be something I could substitute for sin atan x, but I didn't figure out what it should be.

Have you tried to follow whozum's suggestion of drawing a diagram and see how this identity follows:
$$\sin(arctan(x))=\frac{x}{\sqrt{1+x^{2}}}$$
If you haven't, you should try it when you find the time to do so.

 

[Yegor]

I think, that substitution t=1/x isn't more difficult.

 

[arildno]

I think, that substitution t=1/x isn't more difficult.

Also a very good choice, and perhaps the simplest so far.

 

[GCT]

Also a very good choice, and perhaps the simplest so far.

can you clarify? with simplification it seems to lead us right back to trig substitution, or rather a similar form. Perhaps I'm missing something here.

 

[arildno]

Sure:
$$x=\frac{1}{t}\to{dx}=-\frac{dt}{t^{2}}$$
Thus, we have:
$$\int\frac{dx}{x^{2}\sqrt{1+x^{2}}}=-\int\frac{\frac{dt}{t^{2}}}{\frac{1}{t^{2}}\sqrt{1+\frac{1}{t^{2}}}}=-\int\frac{tdt}{\sqrt{1+t^{2}}}=-\sqrt{1+t^{2}}+C=-\frac{\sqrt{x^{2}+1}}{x}+C$$
which is at least as simple as the trig. or hyp. substitutions.