How to integrate this

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fred2028

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How would you go about integrating 2x^3/(x^3 - 1)? Doing long division I get 2 + 2/(x^3 - 1) or similar. Thanks!
 
If we break it into partial fractions, we get:

2∫x3x3−1dx=−23∫xx2+x+1dx−43∫1x2+x+1dx+23∫1x−1dx+2∫dx\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 ∫1x4+1dx\displaystyle \int\frac{1}{x^{4}+1}dx. Perhaps you can work some clever algrbraic way to do yours.

Afterall, it is 2∫1x3−1dx+2∫dx\displaystyle 2\int\frac{1}{x^{3}-1}dx+2\int dx

viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
 
galactus said:
If we break it into partial fractions, we get:

2∫x3x3−1dx=−23∫xx2+x+1dx−43∫1x2+x+1dx+23∫1x−1dx+2∫dx\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 ∫1x4+1dx\displaystyle \int\frac{1}{x^{4}+1}dx. Perhaps you can work some clever algrbraic way to do yours.

Afterall, it is 2∫1x3−1dx+2∫dx\displaystyle 2\int\frac{1}{x^{3}-1}dx+2\int dx

viewtopic.php?f=3&t=28403&p=108823&hilit=+weird#p108823
Click to expand...
How would I integrate x/(x^2 + x + 1)? I can do the other parts, but not this 1. Thanks!
 
fred2028 said:
How would I integrate x/(x^2 + x + 1)? I can do the other parts, but not this 1. Thanks!
Click to expand...

x/(x^2 + x + 1) = (1/2)(2x+1)/(x^2 + x + 1) - (1/2)(1)/(x^2 + x + 1)
 
∫xx2+x+1dx\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.

x2+x+1=(2x+1)24+34\displaystyle x^{2}+x+1=\frac{(2x+1)^{2}}{4}+\frac{3}{4}

Let u=2x+1,   du2=dx,   x=u−12\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:

∫xx2+x+1=12∫⋅2x+1x2+x+1−12∫1x2+x+1\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 u=x2+x+1,   du=(2x+1)dx\displaystyle u=x^{2}+x+1, \;\ du=(2x+1)dx

For the other half, use the sub u=2x+12,   du=dx,   x=2u−12\displaystyle u=\frac{2x+1}{2}, \;\ du=dx, \;\ x=\frac{2u-1}{2}
 
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