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- Improper integral
- Double Integral
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- Curvilinear integral
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How to use it?
- Integral of d{x}:
-
Integral of e^(2*x)/x
-
Integral of ×
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Integral of 1/(x^1/3)
- Integral of (1+x)/x^2
- Graphing y =:
- (x-1)/(x^2-1)
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Identical expressions
- (x- one)/(x^ two - one)
- (x minus 1) divide by (x squared minus 1)
- (x minus one) divide by (x to the power of two minus one)
- (x-1)/(x2-1)
- x-1/x2-1
- (x-1)/(x²-1)
- (x-1)/(x to the power of 2-1)
- x-1/x^2-1
- (x-1) divide by (x^2-1)
- (x-1)/(x^2-1)dx
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Similar expressions
- (x+1)/(x^2-1)
- (cbrt(x))-(1/(x^2-1))
- (x-1)/(x^2+1)
Integral of (x-1)/(x^2-1) dx
The solution
You have entered
[src]
oo / | | x - 1 | ------ dx | 2 | x - 1 | / 2
$$\int\limits_{2}^{\infty} \frac{x - 1}{x^{2} - 1}\, dx$$
Integral((x - 1)/(x^2 - 1), (x, 2, oo))
Detail solution
We have the integral:
/ | | x - 1 | ------ dx | 2 | x - 1 | /
Rewrite the integrand
/ 2*x \
|------------|
| 2 |
x - 1 \x + 0*x - 1/
------ = --------------
2 2
x - 1 or
/ | | x - 1 | ------ dx | 2 = | x - 1 | /
/
|
| 2*x
| ------------ dx
| 2
| x + 0*x - 1
|
/
------------------
2 In the integral
/
|
| 2*x
| ------------ dx
| 2
| x + 0*x - 1
|
/
------------------
2 do replacement
2 u = x
then
the integral =
/
|
| 1
| ------ du
| -1 + u
|
/ log(-1 + u)
------------ = -----------
2 2 do backward replacement
/
|
| 2*x
| ------------ dx
| 2
| x + 0*x - 1
| / 2\
/ log\-1 + x /
------------------ = ------------
2 2 Solution is:
C + log(1 + x)
The answer (Indefinite)
[src]
/ | | x - 1 | ------ dx = C + log(1 + x) | 2 | x - 1 | /
$$\int \frac{x - 1}{x^{2} - 1}\, dx = C + \log{\left(x + 1 \right)}$$
The graph
Use the examples entering the upper and lower limits of integration.