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How to use it?
- Integral of d{x}:
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Integral of (x^2+3)/(x^2-1)
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Integral of (x^3+1)^1/2
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Integral of dx/(2*x+7)
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Integral of 2*x/(x^2+9)
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Identical expressions
- two *x/(x^ two + nine)
- 2 multiply by x divide by (x squared plus 9)
- two multiply by x divide by (x to the power of two plus nine)
- 2*x/(x2+9)
- 2*x/x2+9
- 2*x/(x²+9)
- 2*x/(x to the power of 2+9)
- 2x/(x^2+9)
- 2x/(x2+9)
- 2x/x2+9
- 2x/x^2+9
- 2*x divide by (x^2+9)
- 2*x/(x^2+9)dx
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Similar expressions
- 2*x/(x^2-9)
Integral of 2*x/(x^2+9) dx
The solution
You have entered
[src]
1 / | | 2*x | ------ dx | 2 | x + 9 | / 0
$$\int\limits_{0}^{1} \frac{2 x}{x^{2} + 9}\, dx$$
Integral((2*x)/(x^2 + 9), (x, 0, 1))
Detail solution
We have the integral:
/ | | 2*x | ------ dx | 2 | x + 9 | /
Rewrite the integrand
/0\
|-|
2*x 2*x \9/
------ = ------------ + ----------
2 2 2
x + 9 x + 0*x + 9 /-x \
|---| + 1
\ 3 / or
/ | | 2*x | ------ dx | 2 = | x + 9 | /
/ | | 2*x | ------------ dx | 2 | x + 0*x + 9 | /
In the integral
/ | | 2*x | ------------ dx | 2 | x + 0*x + 9 | /
do replacement
2 u = x
then
the integral =
/ | | 1 | ----- du = log(9 + u) | 9 + u | /
do backward replacement
/ | | 2*x / 2\ | ------------ dx = log\9 + x / | 2 | x + 0*x + 9 | /
In the integral
0
do replacement
-x
v = ---
3 then
the integral =
True
do backward replacement
True
Solution is:
/ 2\ C + log\9 + x /
The answer (Indefinite)
[src]
/ | | 2*x / 2\ | ------ dx = C + log\9 + x / | 2 | x + 9 | /
$$\int \frac{2 x}{x^{2} + 9}\, dx = C + \log{\left(x^{2} + 9 \right)}$$
The graph
The answer
[src]
-log(9) + log(10)
$$- \log{\left(9 \right)} + \log{\left(10 \right)}$$
=
=
-log(9) + log(10)
$$- \log{\left(9 \right)} + \log{\left(10 \right)}$$
-log(9) + log(10)
The graph
Use the examples entering the upper and lower limits of integration.