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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of 1/(2+x^2)
- Integral of x/(x^3-3x+2)
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Integral of -x+4
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Integral of 8/x^2
- How do you in partial fractions?:
- x/(x^3-3x+2)
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Identical expressions
- x/(x^ three -3x+ two)
- x divide by (x cubed minus 3x plus 2)
- x divide by (x to the power of three minus 3x plus two)
- x/(x3-3x+2)
- x/x3-3x+2
- x/(x³-3x+2)
- x/(x to the power of 3-3x+2)
- x/x^3-3x+2
- x divide by (x^3-3x+2)
- x/(x^3-3x+2)dx
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Similar expressions
- x/(x^3+3x+2)
- x/(x^3-3x-2)
Integral of x/(x^3-3x+2) dx
The solution
You have entered
[src]
1 / | | x | ------------ dx | 3 | x - 3*x + 2 | / 0
$$\int\limits_{0}^{1} \frac{x}{\left(x^{3} - 3 x\right) + 2}\, dx$$
Integral(x/(x^3 - 3*x + 2), (x, 0, 1))
The answer (Indefinite)
[src]
/ | | x 2*log(2 + x) 1 2*log(-1 + x) | ------------ dx = C - ------------ - ---------- + ------------- | 3 9 3*(-1 + x) 9 | x - 3*x + 2 | /
$$\int \frac{x}{\left(x^{3} - 3 x\right) + 2}\, dx = C + \frac{2 \log{\left(x - 1 \right)}}{9} - \frac{2 \log{\left(x + 2 \right)}}{9} - \frac{1}{3 \left(x - 1\right)}$$
The answer
[src]
2*pi*I
oo - ------
9
$$\infty - \frac{2 i \pi}{9}$$
=
=
2*pi*I
oo - ------
9
$$\infty - \frac{2 i \pi}{9}$$
oo - 2*pi*i/9
Use the examples entering the upper and lower limits of integration.