Question

How do you use partial fractions to find the integral `int (x^2+x+3)/(x^4+6x^2+9)dx`?

Answer 1

See the answer below:

Answer 2

`1/sqrt(3)tan^(-1)(x/sqrt(3))-1/(2(x^2+3))+C`

Explanation:

Noticing that the denominator is a perfect square, we see there is no need to use partial fractions.

`(x^2+x+3)/(x^4+6x^2+9)=(x^2+3+x)/(x^2+3)^2=cancel(x^2+3)/(x^2+3)^cancel(2)+x/(x^2+3)^2`
`=1/(x^2+3)+x/(x^2+3)^2`
The first term is a standard integral of the form `1/(x^2+a^2)` with `a=sqrt(3)`, yielding `(1/sqrt(3))tan^(-1)(x/sqrt(3))`.

The second term is of the form where the the numerator is the derivative of the contents of the brackets in the denominator, save for a factor of `2`, and so can be integrated at sight. (Alternatively substitute `u=x^2+3`, `dx/(du)=1/(2x)` to give `1/2int u^(-2)du`.)

In general, `d/(dx)(1/(f(x)))=-(f prime(x))/(f(x))^2` so `int (f prime(x))/(f(x))^2dx=-1/(f(x))`. In this case, `f(x)=x^2+3` and `f prime(x)=2x`.