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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/(1+4x^2)
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Integral of (sec(x))^3
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Integral of sec^3
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Integral of x*3^x
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Identical expressions
- x/sqrt(x^ three - one)
- x divide by square root of (x cubed minus 1)
- x divide by square root of (x to the power of three minus one)
- x/√(x^3-1)
- x/sqrt(x3-1)
- x/sqrtx3-1
- x/sqrt(x³-1)
- x/sqrt(x to the power of 3-1)
- x/sqrtx^3-1
- x divide by sqrt(x^3-1)
- x/sqrt(x^3-1)dx
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Similar expressions
- x/sqrt(x^3+1)
Integral of x/sqrt(x^3-1) dx
The solution
You have entered
[src]
1 / | | x | ----------- dx | ________ | / 3 | \/ x - 1 | / 0
$$\int\limits_{0}^{1} \frac{x}{\sqrt{x^{3} - 1}}\, dx$$
Integral(x/(sqrt(x^3 - 1*1)), (x, 0, 1))
The answer (Indefinite)
[src]
_ / 2 |_ /1/2, 2/3 | 3\ | I*x *Gamma(2/3)* | | | x | | x 2 1 \ 5/3 | / | ----------- dx = C - ------------------------------------ | ________ 3*Gamma(5/3) | / 3 | \/ x - 1 | /
$$\int {{{x}\over{\sqrt{x^3-1}}}}{\;dx}$$
The graph
The answer
[src]
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|_ /1/2, 2/3 | \
-I*Gamma(2/3)* | | | 1|
2 1 \ 5/3 | /
----------------------------------
3*Gamma(5/3)
$$\int_{0}^{1}{{{x}\over{\sqrt{x^3-1}}}\;dx}$$
=
=
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|_ /1/2, 2/3 | \
-I*Gamma(2/3)* | | | 1|
2 1 \ 5/3 | /
----------------------------------
3*Gamma(5/3)
$$- \frac{i \Gamma\left(\frac{2}{3}\right) {{}_{2}F_{1}\left(\begin{matrix} \frac{1}{2}, \frac{2}{3} \\ \frac{5}{3} \end{matrix}\middle| {1} \right)}}{3 \Gamma\left(\frac{5}{3}\right)}$$
The graph
Use the examples entering the upper and lower limits of integration.