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-
How to use it?
- Integral of d{x}:
-
Integral of sec^3
-
Integral of sqrt(x)/x
-
Integral of x^2*e^(-x)*dx
-
Integral of (sinx-cosx)^2
- x/(x^2+y^2)
- The double integral of:
- x/(x^2+y^2)
- Canonical form:
- x/(x^2+y^2)
-
Identical expressions
- x/(x^ two +y^ two)
- x divide by (x squared plus y squared )
- x divide by (x to the power of two plus y to the power of two)
- x/(x2+y2)
- x/x2+y2
- x/(x²+y²)
- x/(x to the power of 2+y to the power of 2)
- x/x^2+y^2
- x divide by (x^2+y^2)
- x/(x^2+y^2)dx
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Similar expressions
- x/(x^2-y^2)
Integral of x/(x^2+y^2) dx
The solution
You have entered
[src]
1 / | | x | ------- dx | 2 2 | x + y | / 0
$$\int\limits_{0}^{1} \frac{x}{x^{2} + y^{2}}\, dx$$
Integral(x/(x^2 + y^2), (x, 0, 1))
Detail solution
We have the integral:
/ | | x | ------- dx | 2 2 | x + y | /
Rewrite the integrand
/ 2*x \ /0 \
|-------------| |--|
| 2 2| | 2|
x \x + 0*x + y / \y /
------- = --------------- + ------------
2 2 2 2
x + y /-1 \
|---*x| + 1
\ y / or
/ | | x | ------- dx | 2 2 = | x + y | /
/
|
| 2*x
| ------------- dx
| 2 2
| x + 0*x + y
|
/
-------------------
2 In the integral
/
|
| 2*x
| ------------- dx
| 2 2
| x + 0*x + y
|
/
-------------------
2 do replacement
2 u = x
then
the integral =
/
|
| 1
| ------ du
| 2
| u + y
| / 2\
/ log\u + y /
------------ = -----------
2 2 do backward replacement
/
|
| 2*x
| ------------- dx
| 2 2
| x + 0*x + y
| / 2 2\
/ log\x + y /
------------------- = ------------
2 2 In the integral
0
do replacement
-x
v = ---
y then
the integral =
True
do backward replacement
True
Solution is:
/ 2 2\
log\x + y /
C + ------------
2
The answer (Indefinite)
[src]
/ | / 2 2\ | x log\x + y / | ------- dx = C + ------------ | 2 2 2 | x + y | /
$$\int \frac{x}{x^{2} + y^{2}}\, dx = C + \frac{\log{\left(x^{2} + y^{2} \right)}}{2}$$
The answer
[src]
/ 2\ / 2\
log\1 + y / log\y /
----------- - -------
2 2
$$- \frac{\log{\left(y^{2} \right)}}{2} + \frac{\log{\left(y^{2} + 1 \right)}}{2}$$
=
=
/ 2\ / 2\
log\1 + y / log\y /
----------- - -------
2 2
$$- \frac{\log{\left(y^{2} \right)}}{2} + \frac{\log{\left(y^{2} + 1 \right)}}{2}$$
log(1 + y^2)/2 - log(y^2)/2
Use the examples entering the upper and lower limits of integration.