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- Improper integral
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How to use it?
- Integral of d{x}:
- Integral of x/y
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Integral of exp(2x)
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Integral of dx/(1+e^x)
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Integral of 1/(1+x^2)^1/2
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Identical expressions
- one /(sqrt(x)- one)
- 1 divide by ( square root of (x) minus 1)
- one divide by ( square root of (x) minus one)
- 1/(√(x)-1)
- 1/sqrtx-1
- 1 divide by (sqrt(x)-1)
- 1/(sqrt(x)-1)dx
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Similar expressions
- 1/((sqrtx-1)x^2)
- 1/sqrt((x-1)(2-x))
- (1/sqrtx-1/(x^3)^1/4)
- sqrt(x+1)/sqrt(x-1)
- 1/sqrtx-1/4sqrtx^3
- 1/(sqrt(x)+1)
Integral of 1/(sqrt(x)-1) dx
The solution
You have entered
[src]
1 / | | 1 | 1*--------- dx | ___ | \/ x - 1 | / 0
$$\int\limits_{0}^{1} 1 \cdot \frac{1}{\sqrt{x} - 1}\, dx$$
Integral(1/(sqrt(x) - 1*1), (x, 0, 1))
The answer (Indefinite)
[src]
/ | | 1 ___ / ___\ | 1*--------- dx = C + 2*\/ x + 2*log\-1 + \/ x / | ___ | \/ x - 1 | /
$$2\,\left(\sqrt{x}-1\right)+2\,\log \left(\sqrt{x}-1\right)$$
The answer
[src]
-oo - 2*pi*I
$$\int_{0}^{1}{{{1}\over{\sqrt{x}-1}}\;dx}$$
=
=
-oo - 2*pi*I
$$-\infty - 2 i \pi$$
Use the examples entering the upper and lower limits of integration.