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evaluate the integral

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?

The (1/2) has been missed out, should read:
∫ du/(2u^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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