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- Improper integral
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-
How to use it?
- Integral of d{x}:
- Integral of p/(1-p^2)
-
Integral of 4cos(2x)
-
Integral of 1/cosh(x)
-
Integral of coshx*sinx
-
Identical expressions
- p/(one -p^ two)
- p divide by (1 minus p squared )
- p divide by (one minus p to the power of two)
- p/(1-p2)
- p/1-p2
- p/(1-p²)
- p/(1-p to the power of 2)
- p/1-p^2
- p divide by (1-p^2)
-
Similar expressions
- p/(1+p^2)
Integral of p/(1-p^2) dp
The solution
You have entered
[src]
1 / | | p | ------ dp | 2 | 1 - p | / 0
$$\int\limits_{0}^{1} \frac{p}{- p^{2} + 1}\, dp$$
Detail solution
We have the integral:
/ | | p | 1*------ dp | 2 | 1 - p | /
Rewrite the integrand
-1*2*p + 0
- --------------
2
p - p + 0*p + 1
------ = -----------------
2 2
1 - p or
/ | | p | 1*------ dp | 2 = | 1 - p | /
/
|
| -1*2*p + 0
- | -------------- dp
| 2
| - p + 0*p + 1
|
/
----------------------
2 In the integral
/
|
| -1*2*p + 0
- | -------------- dp
| 2
| - p + 0*p + 1
|
/
----------------------
2 do replacement
2 u = -p
then
the integral =
/
|
| 1
- | ----- du
| 1 + u
|
/ -log(1 + u)
------------- = ------------
2 2 do backward replacement
/
|
| -1*2*p + 0
- | -------------- dp
| 2
| - p + 0*p + 1
| / 2\
/ -log\-1 + p /
---------------------- = --------------
2 2 Solution is:
/ 2\
log\-1 + p /
C - ------------
2
The answer (Indefinite)
[src]
/ | / 2\ | p log\1 - p / | ------ dp = C - ----------- | 2 2 | 1 - p | /
$$-{{\log \left(1-p^2\right)}\over{2}}$$
Use the examples entering the upper and lower limits of integration.