Integral using partial fraction

1. Jul 1, 2010

Telemachus

1. The problem statement, all variables and given/known data

I must solve using partial fractions:

$$\displaystyle\int_{}^{}\displaystyle\frac{dx}{8x^3+1}$$

3. The attempt at a solution

The only real root for the denominator: $$-\displaystyle\frac{1}{2}$$

Then $$8x^3+1=(x+\displaystyle\frac{1}{2})(8x^2-4x+2)$$

Then I did:

$$\displaystyle\frac{1}{(x+\displaystyle\frac{1}{2})(8x^2-4x+2)}=\displaystyle\frac{A}{(x+\displaystyle\frac{1}{2})}+\displaystyle\frac{Bx+C}{(8x^2-4x+2)}$$

I constructed the system:

$$A=\displaystyle\frac{1}{6}$$ $$B=\displaystyle\frac{-4}{3}$$ $$C=\displaystyle\frac{4}{3}$$

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

Completing the square $$8x^2-4x+2=8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}$$

$$\displaystyle\int_{}^{}\displaystyle\frac{x-1}{(8x^2-4x+2)}dx=\displaystyle\int_{}^{}\displaystyle\frac{x-1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx$$

$$\displaystyle\int_{}^{}\displaystyle\frac{x-1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}=\displaystyle\int_{}^{}\displaystyle\frac{x}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx-\displaystyle\int_{}^{}\displaystyle\frac{1}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{2}}dx$$

$$t=x-\displaystyle\frac{1}{4}\Rightarrow{x=t+\displaystyle\frac{1}{4}}$$
$$dt=dx$$

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

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.

Last edited: Jul 1, 2010
2. Jul 1, 2010

Staff: Mentor

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. Jul 1, 2010

Telemachus

Didn't realize of it. Thanks Mark44. Too many fractions really. Its killing me :P

4. Jul 1, 2010

The Chaz

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.