# Question #de09a

Apr 2, 2017

$\frac{1}{4} \ln | x | - \frac{1}{8} \ln | {x}^{2} + 4 | + D$

#### Explanation:

The first step is to make all polynomial fractions into proper fractions, meaning that the degree of the denominator must exceed that of the numerator. But this you already have.

The next step is to factorise the polynomial in the denominator as much as possible. In this case a possible factorisation is
${x}^{3} + 4 x = x \left({x}^{2} + 4\right)$.

If we stick with real numbers, the last factor has no real roots (check this), so we keep it as it is.

Next we do partial fractions, and make the ansatz
$\frac{1}{x \left({x}^{2} + 4\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 4}$,
where $A$,$B$ and $C$ are constants.
(see the link below for why the second factor has $B x + C$ in the denominator).

Multiply both sides by the denominator in the left hand side to get
$1 = A \left({x}^{2} + 4\right) + B {x}^{2} + C x$.
Now for this to hold for any $x$, all the coefficients of the ${x}^{2}$-terms must match on both sides, and similarly for $x$-terms and constant terms.

Then we get a system of equations
$1 = 4 A$
$0 = C x$
$0 = \left(A + B\right) {x}^{2}$.
Solving for the constants, we get that $A = \frac{1}{4}$, $C = 0$, $B = - A = - \frac{1}{4}$.

Thus we have shown that
$\frac{1}{x \left({x}^{2} + 4\right)} = \frac{1}{4 x} - \frac{1}{4} \frac{x}{{x}^{2} + 4}$.

The right hand side we can integrate using standard integrals. The indeterminate integral (unique up to a constant) is then
$\int \frac{1}{4 x} - \frac{1}{4} \frac{x}{{x}^{2} + 4} \mathrm{dx} = \frac{1}{4} \ln | x | - \frac{1}{4} \left(\frac{1}{2} \ln | {x}^{2} + 4 |\right) + D$,
where $D$ is yet another constant. This is the answer.

Note 1: http://www.purplemath.com/modules/partfrac2.htm has a lot of good examples and presents some alternative methods.

Note 2: $\ln \left(\right)$ is the logarithm for base $e$, Euler's number, and $| |$ denotes absolute value.