Integral using partial fraction

[Telemachus]

Homework Statement

I must solve using partial fractions:

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

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.

 

[Mark44]

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.

 

[Telemachus]

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

 

[The Chaz]

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.