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andriy [413]
3 years ago
9

Integral of 1/(x^2-x+1) ​

Mathematics
1 answer:
Lunna [17]3 years ago
6 0

\boxed{\sf \displaystyle{\int}x^ndx=\dfrac{x^{n+1}}{n+1}}

\\ \sf\longmapsto \displaystyle{\int}\dfrac{1}{x^2-x+1}

\\ \sf\longmapsto \displaystyle{\int}x^{-2}-x^{-1}+1

\\ \sf\longmapsto \displaystyle{\int}x^{-2}-\displaystyle{\int}x^{-1}+\displaystyle{\int}1

\\ \sf\longmapsto \dfrac{x^{-2+1}}{-2+1}-\dfrac{x^{-1+1}}{-1+1}+1

\\ \sf\longmapsto \dfrac{x^{-1}}{-1}-\dfrac{x^0}{0}+1

\\ \sf\longmapsto -\dfrac{1}{x}-\dfrac{1}{0}+1

\\ \sf\longmapsto -\dfrac{1}{x}-\infty+1

\\ \sf\longmapsto \infty

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