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