Question

what is the integral of 1/u^2-16

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

table number 20

Answer 6

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

Answer 7

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

Answer 8

(u/4)*

Answer 9

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