Question

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

Answer 1

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

Explanation:

`int5/(4x^3+4)dx = 5/4int1/(x^3+1)dx`

Expand `1/(x^3+1)`:

`int5/(4x^3+4)dx = 5/12int(1/(x+1) - (x -2)/(x^2 - x + 1))dx`

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

Break into two integrals:

`int5/(4x^3+4)dx = 5/12int1/(x+1)dx - 5/12int(x -2)/(x^2 - x + 1)dx`

Multiply the second integral by `2/2`

`int5/(4x^3+4)dx = 5/12int1/(x+1)dx - 5/24int(2x -4)/(x^2 - x + 1)dx`

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

`int5/(4x^3+4)dx = 5/12int1/(x+1)dx - 5/24int(2x -1)/(x^2 - x + 1)dx -5/24int(-3)/(x^2 - x + 1)dx`

Simplify the third integral:

`int5/(4x^3+4)dx = 5/12int1/(x+1)dx - 5/24int(2x -1)/(x^2 - x + 1)dx +5/8int1/(x^2 - x + 1)dx`

The first integral is the natural logarithm:

`int5/(4x^3+4)dx = 5/12ln(x+1) - 5/24int(2x -1)/(x^2 - x + 1)dx +5/8int1/(x^2 - x + 1)dx`

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

`int5/(4x^3+4)dx = 5/12ln(x+1) - 5/24ln(x^2 - x + 1) +5/8int1/(x^2 - x + 1)dx`

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

`int5/(4x^3+4)dx = 5/12ln(x+1) - 5/24ln(x^2 - x + 1) +(5sqrt(3))/12tan^-1((sqrt(3)(2x - 1))/3)+ C`