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How to use it?
- Integral of d{x}:
-
Integral of e^(x*(-5))
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Integral of 1/(x^2-x)
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Integral of (2x-1)(3x+4)
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Integral of 1/cos^3x
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Identical expressions
- (-y)/(two +y^ two)
- ( minus y) divide by (2 plus y squared )
- ( minus y) divide by (two plus y to the power of two)
- (-y)/(2+y2)
- -y/2+y2
- (-y)/(2+y²)
- (-y)/(2+y to the power of 2)
- -y/2+y^2
- (-y) divide by (2+y^2)
-
Similar expressions
- (-y)/(2-y^2)
- (y)/(2+y^2)
Integral of (-y)/(2+y^2) dx
The solution
You have entered
[src]
1 / | | -y | ------ dy | 2 | 2 + y | / 0
$$\int\limits_{0}^{1} \frac{\left(-1\right) y}{y^{2} + 2}\, dy$$
Integral((-y)/(2 + y^2), (y, 0, 1))
Detail solution
We have the integral:
/ | | -y | ------ dy | 2 | 2 + y | /
Rewrite the integrand
/ 2*y \
|------------| /0\
| 2 | |-|
-y \y + 0*y + 2/ \2/
------ = - -------------- + ----------------
2 2 2
2 + y / ___ \
|-\/ 2 |
|-------*y| + 1
\ 2 / or
/ | | -y | ------ dy | 2 = | 2 + y | /
/
|
| 2*y
- | ------------ dy
| 2
| y + 0*y + 2
|
/
--------------------
2 In the integral
/
|
| 2*y
- | ------------ dy
| 2
| y + 0*y + 2
|
/
--------------------
2 do replacement
2 u = y
then
the integral =
/
|
| 1
- | ----- du
| 2 + u
|
/ -log(2 + u)
------------- = ------------
2 2 do backward replacement
/
|
| 2*y
- | ------------ dy
| 2
| y + 0*y + 2
| / 2\
/ -log\2 + y /
-------------------- = -------------
2 2 In the integral
0
do replacement
___
-y*\/ 2
v = ---------
2 then
the integral =
True
do backward replacement
True
Solution is:
/ 2\
log\2 + y /
C - -----------
2
The answer (Indefinite)
[src]
/ | / 2\ | -y log\2 + y / | ------ dy = C - ----------- | 2 2 | 2 + y | /
$$\int \frac{\left(-1\right) y}{y^{2} + 2}\, dy = C - \frac{\log{\left(y^{2} + 2 \right)}}{2}$$
The graph
The answer
[src]
log(2) log(3) ------ - ------ 2 2
$$- \frac{\log{\left(3 \right)}}{2} + \frac{\log{\left(2 \right)}}{2}$$
=
=
log(2) log(3) ------ - ------ 2 2
$$- \frac{\log{\left(3 \right)}}{2} + \frac{\log{\left(2 \right)}}{2}$$
log(2)/2 - log(3)/2
Use the examples entering the upper and lower limits of integration.