Integral using partial fraction

In summary, the student attempted to solve the homework statement using partial fractions, but ran into trouble. He eventually found a simplified solution using the square root of 8x^2-4x+2.
  • #1
Telemachus
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Homework Statement



I must solve using partial fractions:

[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3+1}[/tex]

The Attempt at a Solution



The only real root for the denominator: [tex]-\displaystyle\frac{1}{2}[/tex]

Then [tex]8x^3+1=(x+\displaystyle\frac{1}{2})(8x^2-4x+2)[/tex]

Then I did:

[tex]\displaystyle\frac{1}{(x+\displaystyle\frac{1}{2})(8x^2-4x+2)}=\displaystyle\frac{A}{(x+\displaystyle\frac{1}{2})}+\displaystyle\frac{Bx+C}{(8x^2-4x+2)}[/tex]

I constructed the system:

[tex]A=\displaystyle\frac{1}{6}[/tex] [tex]B=\displaystyle\frac{-4}{3}[/tex] [tex]C=\displaystyle\frac{4}{3}[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3+1}=\displaystyle\frac{1}{6}\displaystyle\int_{}^{}\displaystyle\frac{dx}{(x+\displaystyle\frac{1}{2})}dx-\displaystyle\frac{4}{3}\displaystyle\int_{}^{}\displaystyle\frac{x-1}{(8x^2-4x+2)}dx[/tex]

Completing the square [tex]8x^2-4x+2=8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}[/tex][tex]\displaystyle\int_{}^{}\displaystyle\frac{x-1}{(8x^2-4x+2)}dx=\displaystyle\int_{}^{}\displaystyle\frac{x-1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x-1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}=\displaystyle\int_{}^{}\displaystyle\frac{x}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx-\displaystyle\int_{}^{}\displaystyle\frac{1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx[/tex]

[tex]t=x-\displaystyle\frac{1}{4}\Rightarrow{x=t+\displaystyle\frac{1}{4}}[/tex]
[tex]dt=dx[/tex]

[tex]\displaystyle\int_{}^{}\displaystyle\frac{x}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx=\displaystyle\int_{}^{}\displaystyle\frac{t+\displaystyle\frac{1}{4}}{8t^2+\displaystyle\frac{3}{2}}dt=\displaystyle\frac{1}{8}\displaystyle\int_{}^{}\displaystyle\frac{t}{t^2+\displaystyle\frac{3}{2}}dt+\displaystyle\frac{1}{32}\displaystyle\int_{}^{}\displaystyle\frac{dt}{t^2+\displaystyle\frac{3}{2}}[/tex]

Am I on the right way? Its too tedious. The statement says I must use partial fractions, but anyway if you see a simpler way of solving it let me know.
 
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  • #2
This is the correct approach. You might have made life a little easier on yourself by factoring 8x^3 + 1 into (2x + 1)(4x^2 -2x + 1), thereby eliminating at least some of the fractions.
 
  • #3
Didn't realize of it. Thanks Mark44. Too many fractions really. Its killing me :P
 
  • #4
Telemachus said:
Didn't realize of it. Thanks Mark44. Too many fractions really. Its killing me :P
This is the kind of problem that you shouls do once, for whatever reason (!), and then use a computer for afterward! Good luck. wolframalpha can verify your results, in case you weren't aware of that resource.
 

1. What is partial fraction decomposition?

Partial fraction decomposition is a method used to break down a rational expression into simpler fractions. It involves expressing the rational expression as a sum of simpler fractions with distinct denominators.

2. Why is partial fraction decomposition useful?

Partial fraction decomposition is useful because it allows us to solve integrals of rational functions, which are otherwise difficult to solve. It also helps in simplifying complex algebraic expressions.

3. How do you determine the partial fractions of a rational expression?

To determine the partial fractions of a rational expression, we follow a step-by-step process that involves first factoring the denominator and then setting up a system of equations to solve for the unknown coefficients. This process is known as the method of undetermined coefficients.

4. Are there any restrictions when using partial fraction decomposition?

Yes, there are restrictions when using partial fraction decomposition. The denominator of the rational expression must be factorable into distinct linear or irreducible quadratic factors. Additionally, the degree of the numerator must be less than the degree of the denominator.

5. Can partial fraction decomposition be used for improper integrals?

Yes, partial fraction decomposition can be used for improper integrals. However, it requires additional steps to handle the infinite limits of integration. It is important to check for convergence before using partial fraction decomposition for improper integrals.

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