Question

How do you find the integral of `int 17/(16+x^2)dx`?

Answer 1

`17/4 arctan(x/4)+C`. Standard form: `int1/(a^2+x^2)dx=(1/a)arctan (x/a)+C`

Explanation:

Alternatively substitute `x=4tan u` and use `1+tan^2u=sec^2u` giving the integral as `int (17 xx 4)sec^2 u/(16 sec^2 u)du` and hence the answer.