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answer is 1/8x(log(4-u)-log(4+u)) + constant

\[1/8\left[ \ln \left| u-4/u+4\right|+ C\right]\]

i mean \[1/8[\ln \left|u-4/u+4\right|] + C\]

um, for this integral use would use the tan^-1 form, since in the denominator we have u^2+a^2

table number 20

pull out a -1, then u have int: 1/u^2+16. Then, the solution is (1/4)tan^-1(4u) + C

i mean -(1/4)tan^-1(4u)+C

(u/4)*

it is of the form int_du/u^2-a^2