# How do you integrate f(x)=5/(4x^3+4) using the quotient rule?

Jan 30, 2017

The quotient rule applies to differentiation; not integration. The given function must be integrated using partial fraction expansion, variable substitution, and trigonometric substitution.

#### Explanation:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{4} \int \frac{1}{{x}^{3} + 1} \mathrm{dx}$

Expand $\frac{1}{{x}^{3} + 1}$:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \int \left(\frac{1}{x + 1} - \frac{x - 2}{{x}^{2} - x + 1}\right) \mathrm{dx}$

Note: I would have taken you, step-by-step, through the expansion but it made the explanation too long.

Break into two integrals:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \int \frac{1}{x + 1} \mathrm{dx} - \frac{5}{12} \int \frac{x - 2}{{x}^{2} - x + 1} \mathrm{dx}$

Multiply the second integral by $\frac{2}{2}$

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \int \frac{1}{x + 1} \mathrm{dx} - \frac{5}{24} \int \frac{2 x - 4}{{x}^{2} - x + 1} \mathrm{dx}$

Break the second integral into two integral so that the numerator of the first is $2 x - 1$:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \int \frac{1}{x + 1} \mathrm{dx} - \frac{5}{24} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} - \frac{5}{24} \int \frac{- 3}{{x}^{2} - x + 1} \mathrm{dx}$

Simplify the third integral:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \int \frac{1}{x + 1} \mathrm{dx} - \frac{5}{24} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} + \frac{5}{8} \int \frac{1}{{x}^{2} - x + 1} \mathrm{dx}$

The first integral is the natural logarithm:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \ln \left(x + 1\right) - \frac{5}{24} \int \frac{2 x - 1}{{x}^{2} - x + 1} \mathrm{dx} + \frac{5}{8} \int \frac{1}{{x}^{2} - x + 1} \mathrm{dx}$

A variable substitution makes the second integral become the natural logarithm, too:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \ln \left(x + 1\right) - \frac{5}{24} \ln \left({x}^{2} - x + 1\right) + \frac{5}{8} \int \frac{1}{{x}^{2} - x + 1} \mathrm{dx}$

A trigonometric substitution makes the second integral become the inverse tangent:

$\int \frac{5}{4 {x}^{3} + 4} \mathrm{dx} = \frac{5}{12} \ln \left(x + 1\right) - \frac{5}{24} \ln \left({x}^{2} - x + 1\right) + \frac{5 \sqrt{3}}{12} {\tan}^{-} 1 \left(\frac{\sqrt{3} \left(2 x - 1\right)}{3}\right) + C$