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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/(4+x^2)
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Integral of x+
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Integral of 1/(x^2+4x+8)
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Integral of (1+cos^2x)/(1+cos2x)
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Identical expressions
- one /(x-sqrt(x^ two -x+ one))
- 1 divide by (x minus square root of (x squared minus x plus 1))
- one divide by (x minus square root of (x to the power of two minus x plus one))
- 1/(x-√(x^2-x+1))
- 1/(x-sqrt(x2-x+1))
- 1/x-sqrtx2-x+1
- 1/(x-sqrt(x²-x+1))
- 1/(x-sqrt(x to the power of 2-x+1))
- 1/x-sqrtx^2-x+1
- 1 divide by (x-sqrt(x^2-x+1))
- 1/(x-sqrt(x^2-x+1))dx
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Similar expressions
- 1/(x-sqrt(x^2+x+1))
- 1/(x+sqrt(x^2-x+1))
- 1/(x-sqrt(x^2-x-1))
Integral of 1/(x-sqrt(x^2-x+1)) dx
The solution
You have entered
[src]
1 / | | 1 | 1*------------------- dx | ____________ | / 2 | x - \/ x - x + 1 | / 0
$$\int\limits_{0}^{1} 1 \cdot \frac{1}{x - \sqrt{x^{2} - x + 1}}\, dx$$
Integral(1/(x - sqrt(x^2 - x + 1)), (x, 0, 1))
The answer
[src]
1 / | | 1 | ------------------- dx | ____________ | / 2 | x - \/ 1 + x - x | / 0
$$\int_{0}^{1}{{{1}\over{x-\sqrt{x^2-x+1}}}\;dx}$$
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1 / | | 1 | ------------------- dx | ____________ | / 2 | x - \/ 1 + x - x | / 0
$$\int\limits_{0}^{1} \frac{1}{x - \sqrt{x^{2} - x + 1}}\, dx$$
Use the examples entering the upper and lower limits of integration.