- #1

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Any help would be appreciated, thanks in advance!

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- Thread starter conniebear14
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- #1

- 9

- 0

Any help would be appreciated, thanks in advance!

- #2

arildno

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Furthermore, you have:

du/dx=5u, that is dx=du/(5u)

Negative infinity in "x" goes to 0 in "u", whereas 0 in "x" goes to 1 in "u"

Your new integrand becomes 1/5*(1/(1+u^2)), which you should know how to integrate.

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