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How to use it?
- Integral of d{x}:
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Integral of 0dx
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Integral of x/(3x²-1)^4
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Integral of x^(4/5)
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Integral of (x^2)sinx
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Identical expressions
- one /(x+y+z+ one)^ three
- 1 divide by (x plus y plus z plus 1) cubed
- one divide by (x plus y plus z plus one) to the power of three
- 1/(x+y+z+1)3
- 1/x+y+z+13
- 1/(x+y+z+1)³
- 1/(x+y+z+1) to the power of 3
- 1/x+y+z+1^3
- 1 divide by (x+y+z+1)^3
- 1/(x+y+z+1)^3dx
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Similar expressions
- 1/(x-y+z+1)^3
- 1/(x+y-z+1)^3
- 1/(x+y+z-1)^3
Integral of 1/(x+y+z+1)^3 dx
The solution
You have entered
[src]
x + y / | | 1 | ---------------- dx | 3 | (x + y + z + 1) | / 0
$$\int\limits_{0}^{x + y} \frac{1}{\left(\left(z + \left(x + y\right)\right) + 1\right)^{3}}\, dx$$
Integral(1/((x + y + z + 1)^3), (x, 0, x + y))
The answer
[src]
1 1
----------------------------------- - --------------------------------------------------------------------------
2 2 2 2 2
2 + 2*y + 2*z + 4*y + 4*z + 4*y*z 2 + 2*y + 2*z + 2*(x + y) + 4*y + 4*z + (x + y)*(4 + 4*y + 4*z) + 4*y*z
$$- \frac{1}{2 y^{2} + 4 y z + 4 y + 2 z^{2} + 4 z + 2 \left(x + y\right)^{2} + \left(x + y\right) \left(4 y + 4 z + 4\right) + 2} + \frac{1}{2 y^{2} + 4 y z + 4 y + 2 z^{2} + 4 z + 2}$$
=
=
1 1
----------------------------------- - --------------------------------------------------------------------------
2 2 2 2 2
2 + 2*y + 2*z + 4*y + 4*z + 4*y*z 2 + 2*y + 2*z + 2*(x + y) + 4*y + 4*z + (x + y)*(4 + 4*y + 4*z) + 4*y*z
$$- \frac{1}{2 y^{2} + 4 y z + 4 y + 2 z^{2} + 4 z + 2 \left(x + y\right)^{2} + \left(x + y\right) \left(4 y + 4 z + 4\right) + 2} + \frac{1}{2 y^{2} + 4 y z + 4 y + 2 z^{2} + 4 z + 2}$$
1/(2 + 2*y^2 + 2*z^2 + 4*y + 4*z + 4*y*z) - 1/(2 + 2*y^2 + 2*z^2 + 2*(x + y)^2 + 4*y + 4*z + (x + y)*(4 + 4*y + 4*z) + 4*y*z)
Use the examples entering the upper and lower limits of integration.