- #1

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## Homework Statement

Compute the indefinite integral.

∫(x^2 + 1)^(-5/2) dx

## The Attempt at a Solution

I have a hunch that I need to substitute x = tan(u) but, as always, my lack of trig skills are holding me back.

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- Thread starter flipsvibe
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- #1

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Compute the indefinite integral.

∫(x^2 + 1)^(-5/2) dx

I have a hunch that I need to substitute x = tan(u) but, as always, my lack of trig skills are holding me back.

- #2

Mark44

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Yes, that's the right substition.

If x = tan u, dx = ?

If x = tan u, dx = ?

- #3

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So dx = sec^2 u du?

- #4

Mark44

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When you're done with you trig substitution, your new integral will be completely in terms of u and du, with x and dx completely gone.

- #5

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∫[tan^2(u) + 1]^(-5/2) * sec^2(u) du

==> ∫[sec^2(u)]^(-5/2) * sec^2(u) du

==> ∫[sec^2(u)]^(-3/2) du

after integration I end up with (-3/2)(tanu/secu)

If I continue on and cancel out the cosines, I end up with (-3/2)sinu.

I don't know how to get this back in terms of x.

- #6

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Wait, I think I got this.

I didn't write out all of my steps in my last reply..so I'll go back one or two, to one I skipped.

(-3/2)[tan(u) / √sec^2(u)]

==>(-3/2) [tanu / √(tan^2u + 1)]

==>-3x / 2√(x^2 + 1)

I didn't write out all of my steps in my last reply..so I'll go back one or two, to one I skipped.

(-3/2)[tan(u) / √sec^2(u)]

==>(-3/2) [tanu / √(tan^2u + 1)]

==>-3x / 2√(x^2 + 1)

Last edited:

- #7

Mark44

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Also, don't forget your constant of integration. I didn't check your answer, but you can do that. If your antiderivative is correct, you should be able to differentiate it and get back the original integrand. IOW, d/dx(-3x/2sqrt(x^2 + 1)) should equal (x^2 + 1)^(-5/2).

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