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- Improper integral
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How to use it?
- Integral of d{x}:
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Integral of e^(2*x+1)
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Integral of sin²xcos²x
- Integral of log(1+x)/(1+x^2)
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Integral of (4x-x^2)dx
- The double integral of:
- 1/(x+y)^2
- 1/(x+y)^2
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Identical expressions
- one /(x+y)^ two
- 1 divide by (x plus y) squared
- one divide by (x plus y) to the power of two
- 1/(x+y)2
- 1/x+y2
- 1/(x+y)²
- 1/(x+y) to the power of 2
- 1/x+y^2
- 1 divide by (x+y)^2
- 1/(x+y)^2dx
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Similar expressions
- 1/(x-y)^2
Integral of 1/(x+y)^2 dy
The solution
You have entered
[src]
2 / | | 1 | -------- dy | 2 | (x + y) | / 0
$$\int\limits_{0}^{2} \frac{1}{\left(x + y\right)^{2}}\, dy$$
Integral(1/((x + y)^2), (y, 0, 2))
The answer (Indefinite)
[src]
/ | | 1 1 | -------- dy = C - ----- | 2 x + y | (x + y) | /
$$\int \frac{1}{\left(x + y\right)^{2}}\, dy = C - \frac{1}{x + y}$$
The answer
[src]
1 1 - - ----- x 2 + x
$$- \frac{1}{x + 2} + \frac{1}{x}$$
=
=
1 1 - - ----- x 2 + x
$$- \frac{1}{x + 2} + \frac{1}{x}$$
1/x - 1/(2 + x)
Use the examples entering the upper and lower limits of integration.