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

Question

Christopher Allers · Accepted Answer

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

#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#

Emma Aldridge · Answer

undefined To integrate $ f(x) = \frac{5}{4x^3 + 4} $ using the quotient rule, you first express the function as the quotient of two functions, $ u(x) $ and $ v(x) $. In this case, $ u(x) = 5 $ and $ v(x) = 4x^3 + 4 $. Then, you apply the quotient rule formula:

$$ \int \frac{u(x)}{v(x)} \, dx = \ln|v(x)| + C $$

where $ C $ is the constant of integration. Therefore,

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

To integrate $ \frac{1}{x^3 + 1} $, you can use the substitution method with $ u = x^3 + 1 $, then $ du = 3x^2 \, dx $, and rewrite the integral as:

$$ \frac{1}{3} \int \frac{1}{u} \, du $$

Integrating $ \frac{1}{u} $ with respect to $ u $ gives $ \ln|u| $. Substituting back for $ u $ gives:

$$ \frac{1}{3} \ln|x^3 + 1| + C $$

So, the integral of $ f(x) $ is:

$$ \frac{5}{12} \ln|x^3 + 1| + C $$