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How to use it?
- Integral of d{x}:
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Integral of 1/(x²+1)
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Integral of f(x)=4x
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Integral of x/sin^2(x)
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Integral of x/cosx^2
- Derivative of:
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2*x/(x^2+4)
- Graphing y =:
- 2*x/(x^2+4)
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Identical expressions
- two *x/(x^ two + four)
- 2 multiply by x divide by (x squared plus 4)
- two multiply by x divide by (x to the power of two plus four)
- 2*x/(x2+4)
- 2*x/x2+4
- 2*x/(x²+4)
- 2*x/(x to the power of 2+4)
- 2x/(x^2+4)
- 2x/(x2+4)
- 2x/x2+4
- 2x/x^2+4
- 2*x divide by (x^2+4)
- 2*x/(x^2+4)dx
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Similar expressions
- 2*x/(x^2-4)
Integral of 2*x/(x^2+4) dx
The solution
You have entered
[src]
1 / | | 2*x | ------ dx | 2 | x + 4 | / 0
$$\int\limits_{0}^{1} \frac{2 x}{x^{2} + 4}\, dx$$
Integral((2*x)/(x^2 + 4), (x, 0, 1))
Detail solution
We have the integral:
/ | | 2*x | ------ dx | 2 | x + 4 | /
Rewrite the integrand
/0\
|-|
2*x 2*x \4/
------ = ------------ + ----------
2 2 2
x + 4 x + 0*x + 4 /-x \
|---| + 1
\ 2 / or
/ | | 2*x | ------ dx | 2 = | x + 4 | /
/ | | 2*x | ------------ dx | 2 | x + 0*x + 4 | /
In the integral
/ | | 2*x | ------------ dx | 2 | x + 0*x + 4 | /
do replacement
2 u = x
then
the integral =
/ | | 1 | ----- du = log(4 + u) | 4 + u | /
do backward replacement
/ | | 2*x / 2\ | ------------ dx = log\4 + x / | 2 | x + 0*x + 4 | /
In the integral
0
do replacement
-x
v = ---
2 then
the integral =
True
do backward replacement
True
Solution is:
/ 2\ C + log\4 + x /
The answer (Indefinite)
[src]
/ | | 2*x / 2\ | ------ dx = C + log\4 + x / | 2 | x + 4 | /
$$\int \frac{2 x}{x^{2} + 4}\, dx = C + \log{\left(x^{2} + 4 \right)}$$
The graph
The answer
[src]
-log(4) + log(5)
$$- \log{\left(4 \right)} + \log{\left(5 \right)}$$
=
=
-log(4) + log(5)
$$- \log{\left(4 \right)} + \log{\left(5 \right)}$$
-log(4) + log(5)
The graph
Use the examples entering the upper and lower limits of integration.