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[tex]\int \frac{dx}{x^2 \sqrt{1+x^2}}[/tex]

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

[itex]u = (1+x^2)^{-1/2}[/itex]

[itex]v = -x^{-1}[/itex]

Then

[itex]u' = -x(1+x^2)^{-3/2}[/itex]

[itex]v' = x^{-2}[/itex]

And so the original problem is equivalent to:

[tex]\int uv' = uv - \int u'v[/tex]

[tex]= -\frac {1}{x (1+x^2)^{1/2}} - \int \frac{(-x)(-x^{-1}) dx}{(1+x^2)^{3/2}}[/tex]

[tex]= -\frac {1}{x (1+x^2)^{1/2}} - \int \frac{dx}{(1+x^2)^{3/2}}[/tex]

... which calls for this trigonometric substitution:

tan y = x

[tex]\frac{dy}{\cos^2 y} = dx[/tex]

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

[tex]-\frac {1}{x (1+x^2)^{1/2}} - \int \frac{dy}{\cos^2 y(\frac{1}{\cos^2 y})^{3/2}}[/tex]

[tex]= -\frac {1}{x (1+x^2)^{1/2}} - \int \cos y \; dy[/tex]

which is

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

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?

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# Homework Help: Antiderivative for 1/[x²√(1+x²)]?

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