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How to integrate this
- Thread starter fred2028
- Start date
If we break it into partial fractions, we get:
\(\displaystyle 2\int\frac{x^{3}}{x^{3}-1}dx=\frac{-2}{3}\int\frac{x}{x^{2}+x+1}dx-\frac{4}{3}\int\frac{1}{x^{2}+x+1}dx+\frac{2}{3}\int\frac{1}{x-1}dx+2\int dx\)
That is one way to proceed. Can you?.
Here is a clever way of doing \(\displaystyle \int\frac{1}{x^{4}+1}dx\). Perhaps you can work some clever algrbraic way to do yours.
Afterall, it is \(\displaystyle 2\int\frac{1}{x^{3}-1}dx+2\int dx\)
viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
\(\displaystyle 2\int\frac{x^{3}}{x^{3}-1}dx=\frac{-2}{3}\int\frac{x}{x^{2}+x+1}dx-\frac{4}{3}\int\frac{1}{x^{2}+x+1}dx+\frac{2}{3}\int\frac{1}{x-1}dx+2\int dx\)
That is one way to proceed. Can you?.
Here is a clever way of doing \(\displaystyle \int\frac{1}{x^{4}+1}dx\). Perhaps you can work some clever algrbraic way to do yours.
Afterall, it is \(\displaystyle 2\int\frac{1}{x^{3}-1}dx+2\int dx\)
viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
How would I integrate x/(x^2 + x + 1)? I can do the other parts, but not this 1. Thanks!galactus said:If we break it into partial fractions, we get:
\(\displaystyle 2\int\frac{x^{3}}{x^{3}-1}dx=\frac{-2}{3}\int\frac{x}{x^{2}+x+1}dx-\frac{4}{3}\int\frac{1}{x^{2}+x+1}dx+\frac{2}{3}\int\frac{1}{x-1}dx+2\int dx\)
That is one way to proceed. Can you?.
Here is a clever way of doing \(\displaystyle \int\frac{1}{x^{4}+1}dx\). Perhaps you can work some clever algrbraic way to do yours.
Afterall, it is \(\displaystyle 2\int\frac{1}{x^{3}-1}dx+2\int dx\)
viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
D
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fred2028 said:How would I integrate x/(x^2 + x + 1)? I can do the other parts, but not this 1. Thanks!
x/(x^2 + x + 1) = (1/2)(2x+1)/(x^2 + x + 1) - (1/2)(1)/(x^2 + x + 1)
\(\displaystyle \int\frac{x}{x^{2}+x+1}dx\)
When you have a quadratic in the denominator that does not factor to well, try completing the square and making what is in the parentheses the u sub.
\(\displaystyle x^{2}+x+1=\frac{(2x+1)^{2}}{4}+\frac{3}{4}\)
Let \(\displaystyle u=2x+1, \;\ \frac{du}{2}=dx, \;\ x=\frac{u-1}{2}\)
Also, you can still break that one up into a partial fraction again:
\(\displaystyle \int\frac{x}{x^{2}+x+1}=\frac{1}{2}\int\cdot\frac{2x+1}{x^{2}+x+1}-\frac{1}{2}\int\frac{1}{x^{2}+x+1}\)
The first half is easily done by letting \(\displaystyle u=x^{2}+x+1, \;\ du=(2x+1)dx\)
For the other half, use the sub \(\displaystyle u=\frac{2x+1}{2}, \;\ du=dx, \;\ x=\frac{2u-1}{2}\)
When you have a quadratic in the denominator that does not factor to well, try completing the square and making what is in the parentheses the u sub.
\(\displaystyle x^{2}+x+1=\frac{(2x+1)^{2}}{4}+\frac{3}{4}\)
Let \(\displaystyle u=2x+1, \;\ \frac{du}{2}=dx, \;\ x=\frac{u-1}{2}\)
Also, you can still break that one up into a partial fraction again:
\(\displaystyle \int\frac{x}{x^{2}+x+1}=\frac{1}{2}\int\cdot\frac{2x+1}{x^{2}+x+1}-\frac{1}{2}\int\frac{1}{x^{2}+x+1}\)
The first half is easily done by letting \(\displaystyle u=x^{2}+x+1, \;\ du=(2x+1)dx\)
For the other half, use the sub \(\displaystyle u=\frac{2x+1}{2}, \;\ du=dx, \;\ x=\frac{2u-1}{2}\)