# How do you integrate 1/(x^3 +4x) using partial fractions?

Nov 13, 2017

The answer is $= \frac{1}{4} \ln \left(| x |\right) - \frac{1}{8} \ln \left({x}^{2} + 4\right) + C$

#### Explanation:

Perform the decomposition into partial fractions

$\frac{1}{{x}^{3} + 4 x} = \frac{1}{x \left({x}^{2} + 4\right)} = \frac{A}{x} + \frac{B x + C}{{x}^{2} + 4}$

$= \frac{A \left({x}^{2} + 4\right) + x \left(B x + C\right)}{x \left({x}^{2} + 4\right)}$

The denominators are the same, compare the numerators

$1 = A \left({x}^{2} + 4\right) + x \left(B x + C\right)$

Let, $x = 0$, $\implies$, $1 = 4 A$, $\implies$, $A = \frac{1}{4}$

Coefficients of ${x}^{2}$

$0 = A + B$, $\implies$, $B = - A = - \frac{1}{4}$

Coefficients of $x$

$0 = C$

Therefore,

$\frac{1}{{x}^{3} + 4 x} = \frac{\frac{1}{4}}{x} + \frac{\left(- \frac{1}{4}\right) x}{{x}^{2} + 4}$

So,

$\int \frac{\mathrm{dx}}{{x}^{3} + 4 x} = \int \frac{\frac{1}{4} \mathrm{dx}}{x} + \int \frac{\left(- \frac{1}{4} \mathrm{dx}\right) x}{{x}^{2} + 4}$

$= \frac{1}{4} \int \frac{\mathrm{dx}}{x} - \frac{1}{4} \int \frac{x \mathrm{dx}}{{x}^{2} + 4}$

Perform the second integral by substitution

Let $u = {x}^{2} + 4$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$

So

$\frac{1}{4} \int \frac{x \mathrm{dx}}{{x}^{2} + 4} = \frac{1}{8} \int \frac{\mathrm{du}}{u} = \frac{1}{8} \ln u = \frac{1}{8} \ln \left({x}^{2} + 4\right)$

$\int \frac{\mathrm{dx}}{{x}^{3} + 4 x} = \frac{1}{4} \ln \left(| x |\right) - \frac{1}{8} \ln \left({x}^{2} + 4\right) + C$

Nov 13, 2017

$\frac{1}{4} \cdot L n x - \frac{1}{8} \cdot L n \left({x}^{2} + 4\right) + C = \frac{1}{8} \cdot L n \left({x}^{2} / \left({x}^{2} + 4\right)\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{{x}^{3} + 4 x}$

=$\int \frac{\mathrm{dx}}{x \cdot \left({x}^{2} + 4\right)}$

=$\frac{1}{4} \cdot \int \frac{4 \cdot \mathrm{dx}}{x \cdot \left({x}^{2} + 4\right)}$

=$\frac{1}{4} \cdot \int \frac{\left({x}^{2} + 4\right) \cdot \mathrm{dx}}{x \left({x}^{2} + 4\right)}$-$\frac{1}{4} \cdot \int \frac{{x}^{2} \cdot \mathrm{dx}}{x \left({x}^{2} + 4\right)}$

=$\frac{1}{4} \cdot \int \frac{\mathrm{dx}}{x}$-$\frac{1}{4} \cdot \int \frac{x \cdot \mathrm{dx}}{{x}^{2} + 4}$

=$\frac{1}{4} \cdot L n x - \frac{1}{8} \cdot L n \left({x}^{2} + 4\right) + C$

=$\frac{1}{8} \cdot L n \left({x}^{2} / \left({x}^{2} + 4\right)\right) + C$

Nov 13, 2017

$\frac{1}{8} \ln | {x}^{2} / \left({x}^{2} + 4\right) | + C .$

#### Explanation:

Here is another way to solve the Problem, but without the

use of Partial Fractions.

Let, $I = \int \frac{1}{{x}^{3} + 4 x} \mathrm{dx} = \int \frac{1}{x \left({x}^{2} + 4\right)} \mathrm{dx} = \int \frac{x}{{x}^{2} \left({x}^{2} + 4\right)} \mathrm{dx} ,$ or,

$I = \frac{1}{2} \int \frac{2 x}{{x}^{2} \left({x}^{2} + 4\right)} \mathrm{dx} .$

Subst. ${x}^{2} = t \Rightarrow 2 x \mathrm{dx} = \mathrm{dt} .$

$\therefore I = \frac{1}{2} \int \frac{1}{t \left(t + 4\right)} \mathrm{dt} = \frac{1}{2} \int \frac{1}{{t}^{2} + 4 t} \mathrm{dt} ,$

$= \frac{1}{2} \int \frac{1}{\left({t}^{2} + 4 t + 4 - 4\right)} \mathrm{dt} ,$

$= \frac{1}{2} \int \frac{1}{{\left(t + 2\right)}^{2} - {2}^{2}} \mathrm{dt} .$

Since, $\int \frac{1}{{y}^{2} - {a}^{2}} \mathrm{dy} = \frac{1}{2 a} \ln | \frac{y - a}{y + a} | ,$

$I = \frac{1}{2} \cdot \frac{1}{2 \cdot 2} \ln | \frac{\left(t + 2\right) - 2}{\left(t + 2\right) + 2} | ,$

$= \frac{1}{8} \ln | \frac{t}{t + 4} | ,$

$\Rightarrow I = \frac{1}{8} \ln | {x}^{2} / \left({x}^{2} + 4\right) | + C .$