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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^3
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Integral of -cos(2x)
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Integral of dx/(4x-1)
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Integral of 2lnx
- The double integral of:
- 1/(x+y+1)^2
- 1/(x+y+1)^2
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Identical expressions
- one /(x+y+ one)^ two
- 1 divide by (x plus y plus 1) squared
- one divide by (x plus y plus one) to the power of two
- 1/(x+y+1)2
- 1/x+y+12
- 1/(x+y+1)²
- 1/(x+y+1) to the power of 2
- 1/x+y+1^2
- 1 divide by (x+y+1)^2
- 1/(x+y+1)^2dx
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Similar expressions
- 1/(x-y+1)^2
- 1/(x+y-1)^2
Integral of 1/(x+y+1)^2 dy
The solution
You have entered
[src]
0 / | | 1 | ------------ dy | 2 | (x + y + 1) | / 1
$$\int\limits_{1}^{0} \frac{1}{\left(\left(x + y\right) + 1\right)^{2}}\, dy$$
Integral(1/((x + y + 1)^2), (y, 1, 0))
The answer (Indefinite)
[src]
/ | | 1 1 | ------------ dy = C - --------- | 2 1 + x + y | (x + y + 1) | /
$$\int \frac{1}{\left(\left(x + y\right) + 1\right)^{2}}\, dy = C - \frac{1}{x + y + 1}$$
The answer
[src]
1 1 ----- - ----- 2 + x 1 + x
$$\frac{1}{x + 2} - \frac{1}{x + 1}$$
=
=
1 1 ----- - ----- 2 + x 1 + x
$$\frac{1}{x + 2} - \frac{1}{x + 1}$$
1/(2 + x) - 1/(1 + x)
Use the examples entering the upper and lower limits of integration.