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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of 2*x/(x^2+1)
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Integral of tan
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Integral of ln(2x)
- Integral of sin6x
- Graphing y =:
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2*x/(x^2+1)
- Derivative of:
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2*x/(x^2+1)
- How do you in partial fractions?:
- 2*x/(x^2+1)
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Identical expressions
- two *x/(x^ two + one)
- 2 multiply by x divide by (x squared plus 1)
- two multiply by x divide by (x to the power of two plus one)
- 2*x/(x2+1)
- 2*x/x2+1
- 2*x/(x²+1)
- 2*x/(x to the power of 2+1)
- 2x/(x^2+1)
- 2x/(x2+1)
- 2x/x2+1
- 2x/x^2+1
- 2*x divide by (x^2+1)
- 2*x/(x^2+1)dx
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Similar expressions
- 2*x/(x^2-1)
Integral of 2*x/(x^2+1) dx
The solution
You have entered
[src]
1 / | | 2*x | ------ dx | 2 | x + 1 | / 0
$$\int\limits_{0}^{1} \frac{2 x}{x^{2} + 1}\, dx$$
Integral((2*x)/(x^2 + 1), (x, 0, 1))
Detail solution
We have the integral:
/ | | 2*x | ------ dx | 2 | x + 1 | /
Rewrite the integrand
/0\
|-|
2*x 2*x \1/
------ = ------------ + ---------
2 2 2
x + 1 x + 0*x + 1 (-x) + 1or
/ | | 2*x | ------ dx | 2 = | x + 1 | /
/ | | 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 | x + 1 | /
$$\int \frac{2 x}{x^{2} + 1}\, dx = C + \log{\left(x^{2} + 1 \right)}$$
The graph
The graph
Use the examples entering the upper and lower limits of integration.