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- Improper integral
- Double Integral
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How to use it?
- Integral of d{x}:
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Integral of e^(2*x+1)
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Integral of (1+x)/x
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Integral of 1/5x
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Integral of (1+2x^2)/(x^2(1+x^2))
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Identical expressions
- exp(x)/(x^ two - one)
- exponent of (x) divide by (x squared minus 1)
- exponent of (x) divide by (x to the power of two minus one)
- exp(x)/(x2-1)
- expx/x2-1
- exp(x)/(x²-1)
- exp(x)/(x to the power of 2-1)
- expx/x^2-1
- exp(x) divide by (x^2-1)
- exp(x)/(x^2-1)dx
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Similar expressions
- exp(x)/(x^2+1)
Integral of exp(x)/(x^2-1) dx
The solution
You have entered
[src]
4 / | | x | e | ------ dx | 2 | x - 1 | / 0
$$\int\limits_{0}^{4} \frac{e^{x}}{x^{2} - 1}\, dx$$
Integral(exp(x)/(x^2 - 1), (x, 0, 4))
The answer (Indefinite)
[src]
/ / | | | x | x | e | e | ------ dx = C + | ---------------- dx | 2 | (1 + x)*(-1 + x) | x - 1 | | / /
$$\int \frac{e^{x}}{x^{2} - 1}\, dx = C + \int \frac{e^{x}}{\left(x - 1\right) \left(x + 1\right)}\, dx$$
The answer
[src]
4 / | | x | e | ---------------- dx | (1 + x)*(-1 + x) | / 0
$$\int\limits_{0}^{4} \frac{e^{x}}{\left(x - 1\right) \left(x + 1\right)}\, dx$$
=
=
4 / | | x | e | ---------------- dx | (1 + x)*(-1 + x) | / 0
$$\int\limits_{0}^{4} \frac{e^{x}}{\left(x - 1\right) \left(x + 1\right)}\, dx$$
Integral(exp(x)/((1 + x)*(-1 + x)), (x, 0, 4))
Use the examples entering the upper and lower limits of integration.