## anonymous 5 years ago what is the integral of 1/u^2-16

1. anonymous

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

table number 20

6. anonymous

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

7. anonymous

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

8. anonymous

(u/4)*

9. anonymous

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