Solving an integral using partial fractions

• July 1st 2010, 04:13 PM
Ulysses
Solving an integral using partial fractions
I must solve using partial fractions:

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

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:

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

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

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

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

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

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

$\displaystyle\int_{}^{}\displaystyle\frac{x}{8(x-\displaystyle\frac{1}{4})^2+\displaystyle\frac{3}{ 2}}dx=\displaystyle\int_{}^{}\displaystyle\frac{t+ \displaystyle\frac{1}{4}}{8t^2+\displaystyle\frac{ 3}{2}}dt=\displaystyle\frac{1}{8}\displaystyle\int _{}^{}\displaystyle\frac{t}{t^2+\displaystyle\frac {3}{2}}dt+\displaystyle\frac{1}{32}\displaystyle\i nt_{}^{}\displaystyle\frac{dt}{t^2+\displaystyle\f rac{3}{2}}$

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.
• July 1st 2010, 05:26 PM
sa-ri-ga-ma
$\frac{4}{3}\int \frac{x-1}{(8x^2 - 4x + 2)} dx$

can be written as

$\frac{2}{3}\int \frac{x-1}{(4x^2 - 2x + 1)}dx$

The numerator can be written as 1/8(8x - 2 - 6)

$\frac{2}{3\times 8}[\int \frac{8x-2}{(4x^2 - 2x + 1)}dx - \int \frac{6}{(4x^2 - 2x + 1)}dx][$

The first integration is in form of f'(x)/f(x), which is equal to log[f(x)] and second integration can be solved by complting the square in the dinominator.
• July 1st 2010, 05:31 PM
mr fantastic
Quote:

Originally Posted by Ulysses
I must solve using partial fractions:

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

[snip]

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.

There is no avoiding tedious algebra.

See here: integrate 1&#47;&#40;8x&#94;3 &#43; 1&#41; - Wolfram|Alpha

and click on Show steps.