integral( (ln(x^2+x+1)) / x^2 ) dx
- Thread starter cheffy
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pka
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Use this link: http://integrals.wolfram.com/index.jsp
Find the antiderivative. Then differentiate to see how it is done.
Find the antiderivative. Then differentiate to see how it is done.
Here's an idea, though I will leave the hard work up to you.
Use parts in the beginning:
\(\displaystyle \L\\\int\frac{ln(x^{2}+x+1)}{x^{2}}dx\)
Let \(\displaystyle \L\\u=ln(x^{2}+x+1), \;\ dv=\frac{1}{x^{2}}dx, \;\ du=\frac{2x+1}{x^{2}+x+1}dx, \;\ v=\frac{-1}{x}\)
Putting 'er all together we get:
\(\displaystyle \L\\\frac{-ln(x^{2}+x+1)}{x}-\int\frac{-2x-1}{x^{3}+x^{2}+x}dx\)
You can use partial fractions on the right side:
\(\displaystyle \L\\\frac{-2x-1}{x^{2}+x^{2}+x}=\frac{x}{x^{2}+x+1}-\frac{1}{x^{2}+x+1}-\frac{1}{x}\)
Good luck.
Use parts in the beginning:
\(\displaystyle \L\\\int\frac{ln(x^{2}+x+1)}{x^{2}}dx\)
Let \(\displaystyle \L\\u=ln(x^{2}+x+1), \;\ dv=\frac{1}{x^{2}}dx, \;\ du=\frac{2x+1}{x^{2}+x+1}dx, \;\ v=\frac{-1}{x}\)
Putting 'er all together we get:
\(\displaystyle \L\\\frac{-ln(x^{2}+x+1)}{x}-\int\frac{-2x-1}{x^{3}+x^{2}+x}dx\)
You can use partial fractions on the right side:
\(\displaystyle \L\\\frac{-2x-1}{x^{2}+x^{2}+x}=\frac{x}{x^{2}+x+1}-\frac{1}{x^{2}+x+1}-\frac{1}{x}\)
Good luck.
Count Iblis
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- Feb 13, 2007
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cheffy said:
Substitute x = 1/y. The integral becomes
\(\displaystyle \int\left[2\log\left(y\right)-\log\left(1+y+y^{2}\right)\right]dy\)
Calculate the second term using a partial integration.
Let's step through this. It's a rather laborious integral. I am sure there are other ways. Someone else may post some clever show-off technique, but here goes.
We join our story already in progress:
\(\displaystyle \L\\\int\frac{-2x-1}{x^{3}+x^{2}+x}dx\)
\(\displaystyle \L\\\left(\frac{-1}{x}+\frac{x-1}{x^{2}+x+1}\right)dx\)
\(\displaystyle \L\\\int\frac{-1}{x}dx+\int\frac{x-1}{x^{2}+x+1}dx\)
\(\displaystyle \L\\-ln(x)+\int\frac{x-1}{x^{2}+x+1}dx\)
\(\displaystyle \L\\-ln(x)+\underbrace{\int\frac{1}{2}\frac{2x+1}{x^{2}+x+1}dx}_{\text{[1]}}-\underbrace{\frac{3}{2}\int\frac{1}{x^{2}+x+1}dx}_{\text{[2]}}\)
For [1], let \(\displaystyle u=x^{2}+x+1, \;\ du=(2x+1)dx\)
For [1], this gives:
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}ln(u)=\frac{1}{2}ln(x^{2}+x+1)\)
Now for [2]:
\(\displaystyle \L\\\frac{3}{2}\int\frac{1}{x^{2}+x+1}dx\)
Let \(\displaystyle u=\frac{2x+1}{2}, \;\ du=dx, \;\ x=\frac{2u-1}{2}\)
\(\displaystyle \L\\\frac{3}{2}\int\frac{4}{4u^{2}+3}du\)
\(\displaystyle \L\\\sqrt{3}tan^{-1}(\frac{2\sqrt{3}u}{3})\)
=\(\displaystyle \L\\\sqrt{3}tan^{-1}(\frac{\sqrt{3}(2x+1)}{3})}\)
Now, put this all together with the other parts and we have:
\(\displaystyle \H\\\frac{-ln(x^{2}+x+1)}{x}+ln(x)-\frac{ln(x^{2}+x+1)}{2}+\sqrt{3}tan^{-1}\left(\frac{\sqrt{3}(2x+1)}{3}\right)\)
We join our story already in progress:
\(\displaystyle \L\\\int\frac{-2x-1}{x^{3}+x^{2}+x}dx\)
\(\displaystyle \L\\\left(\frac{-1}{x}+\frac{x-1}{x^{2}+x+1}\right)dx\)
\(\displaystyle \L\\\int\frac{-1}{x}dx+\int\frac{x-1}{x^{2}+x+1}dx\)
\(\displaystyle \L\\-ln(x)+\int\frac{x-1}{x^{2}+x+1}dx\)
\(\displaystyle \L\\-ln(x)+\underbrace{\int\frac{1}{2}\frac{2x+1}{x^{2}+x+1}dx}_{\text{[1]}}-\underbrace{\frac{3}{2}\int\frac{1}{x^{2}+x+1}dx}_{\text{[2]}}\)
For [1], let \(\displaystyle u=x^{2}+x+1, \;\ du=(2x+1)dx\)
For [1], this gives:
\(\displaystyle \L\\\frac{1}{2}\int\frac{1}{u}du=\frac{1}{2}ln(u)=\frac{1}{2}ln(x^{2}+x+1)\)
Now for [2]:
\(\displaystyle \L\\\frac{3}{2}\int\frac{1}{x^{2}+x+1}dx\)
Let \(\displaystyle u=\frac{2x+1}{2}, \;\ du=dx, \;\ x=\frac{2u-1}{2}\)
\(\displaystyle \L\\\frac{3}{2}\int\frac{4}{4u^{2}+3}du\)
\(\displaystyle \L\\\sqrt{3}tan^{-1}(\frac{2\sqrt{3}u}{3})\)
=\(\displaystyle \L\\\sqrt{3}tan^{-1}(\frac{\sqrt{3}(2x+1)}{3})}\)
Now, put this all together with the other parts and we have:
\(\displaystyle \H\\\frac{-ln(x^{2}+x+1)}{x}+ln(x)-\frac{ln(x^{2}+x+1)}{2}+\sqrt{3}tan^{-1}\left(\frac{\sqrt{3}(2x+1)}{3}\right)\)