- #1
Telemachus
- 835
- 30
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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