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- Improper integral
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-
How to use it?
- Integral of d{x}:
-
Integral of 1/sqrt(x^2+1)
-
Integral of x^3*ln(x)
-
Integral of x^2*lnx
- Integral of 1/(x(1+x^2))
- Graphing y =:
- (x+1)/(x^2-1)
-
Identical expressions
- (x+ one)/(x^ two - one)
- (x plus 1) divide by (x squared minus 1)
- (x plus 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)
- (x+1)/(x^2+1)
Integral of (x+1)/(x^2-1) dx
The solution
You have entered
[src]
1 / | | x + 1 | ------ dx | 2 | x - 1 | / 0
$$\int\limits_{0}^{1} \frac{x + 1}{x^{2} - 1}\, dx$$
Integral((x + 1)/(x^2 - 1*1), (x, 0, 1))
Detail solution
We have the integral:
/ | | x + 1 | 1*------ dx | 2 | x - 1 | /
Rewrite the integrand
/ 1*2*x + 0 \
|--------------|
| 2 |
x + 1 \1*x + 0*x - 1/
------ = ----------------
2 2
x - 1 or
/ | | x + 1 | 1*------ dx | 2 = | x - 1 | /
/
|
| 1*2*x + 0
| -------------- dx
| 2
| 1*x + 0*x - 1
|
/
--------------------
2 In the integral
/
|
| 1*2*x + 0
| -------------- dx
| 2
| 1*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
/
|
| 1*2*x + 0
| -------------- dx
| 2
| 1*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 | /
$$\log \left(x-1\right)$$
Use the examples entering the upper and lower limits of integration.