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March 27, 2015

March 27, 2015

Posted by **Hannah** on Saturday, April 30, 2011 at 6:06pm.

integral of 3 to 2 x/(x^2-2)^2 dx

u=x^2-2

du=2x dx

1/2 du = x dx

integral of 1/u^2 du

-1/(x^2-2) Then I plug in 3 and 2 and subtract them form each other

-1/(3^2-2) - (-1/(2^2-2) Is this correct?

- Math(Please check) -
**MathMate**, Saturday, April 30, 2011 at 9:09pmThe (1/2) has been missed out, should read:

∫ du/(**2**u^2)

=-1/(**2**(x^2-2))

After that you can plug in the limits of integration. I get 5/28 integrating from 2 to 3.

The bottom limit is the start value, and the top value is the end value.

It is usual to go from 2 to 3, and rarely from 3 to 2. Please check.

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