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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of e^(4*x)*dx
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Integral of 1/(x^2-9)
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Integral of 2*x/(-1+x)
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Integral of 1/(2+3x^2)
- Limit of the function:
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x/(1-3*x^2)
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Identical expressions
- x/(one - three *x^ two)
- x divide by (1 minus 3 multiply by x squared )
- x divide by (one minus three multiply by x to the power of two)
- x/(1-3*x2)
- x/1-3*x2
- x/(1-3*x²)
- x/(1-3*x to the power of 2)
- x/(1-3x^2)
- x/(1-3x2)
- x/1-3x2
- x/1-3x^2
- x divide by (1-3*x^2)
- x/(1-3*x^2)dx
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Similar expressions
- x/(1+3*x^2)
Integral of x/(1-3*x^2) dx
The solution
You have entered
[src]
1 / | | x | -------- dx | 2 | 1 - 3*x | / 0
$$\int\limits_{0}^{1} \frac{x}{1 - 3 x^{2}}\, dx$$
Integral(x/(1 - 3*x^2), (x, 0, 1))
Detail solution
We have the integral:
/ | | x | -------- dx | 2 | 1 - 3*x | /
Rewrite the integrand
-3*2*x
- ----------------
2
x - 3*x + 0*x + 1
-------- = -------------------
2 6
1 - 3*x or
/ | | x | -------- dx | 2 = | 1 - 3*x | /
/
|
| -3*2*x
- | ---------------- dx
| 2
| - 3*x + 0*x + 1
|
/
------------------------
6 In the integral
/
|
| -3*2*x
- | ---------------- dx
| 2
| - 3*x + 0*x + 1
|
/
------------------------
6 do replacement
2 u = -3*x
then
the integral =
/
|
| 1
- | ----- du
| 1 + u
|
/ -log(1 + u)
------------- = ------------
6 6 do backward replacement
/
|
| -3*2*x
- | ---------------- dx
| 2
| - 3*x + 0*x + 1
| / 2\
/ -log\-1 + 3*x /
------------------------ = ----------------
6 6 Solution is:
/ 2\
log\-1 + 3*x /
C - --------------
6
The answer (Indefinite)
[src]
/ | / 2\ | x log\1 - 3*x / | -------- dx = C - ------------- | 2 6 | 1 - 3*x | /
$$\int \frac{x}{1 - 3 x^{2}}\, dx = C - \frac{\log{\left(1 - 3 x^{2} \right)}}{6}$$
The graph
Use the examples entering the upper and lower limits of integration.