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How to use it?
How do you in partial fractions?
:
2*a*(3*a-p)/(2*p-6*a)
2*x/(x^2+1)-2*x*(x^2-1)/(x^2+1)^2
(((y-y)/(3*y-3))+(1/(y-1)))/((y+1)/3)+(2/(y^2-1))
(x^2-12)/(x-3)-x/(3-x)
Factor polynomial
:
x^3-1
x^7+1
c^5+1
x^2-4
Least common denominator
:
(z/b-b/z)*6*z*b/(z+b)
(((z)/(b))-((b)/(z)))*((4*z*b)/(z+b))
(z/8+z/3)/z^2
(z-6*z/z+3)/z-3/z+3
Factor squared
:
y^4+y^2-1
-y^4+9*y^2+5
y^4-9*y^2+4
x^2*(-2)+5*x-2
Integral of d{x}
:
(x+1)/(x^2+1)
Derivative of
:
(x+1)/(x^2+1)
Graphing y =
:
(x+1)/(x^2+1)
Identical expressions
(x+ one)/(x^ two + one)
(x plus 1) divide by (x squared plus 1)
(x plus one) divide by (x to the power of two plus one)
(x+1)/(x2+1)
x+1/x2+1
(x+1)/(x²+1)
(x+1)/(x to the power of 2+1)
x+1/x^2+1
(x+1) divide by (x^2+1)
Similar expressions
(x+1)/(x^2-1)
(x-1)/(x^2+1)
Expression simplification
/
Fraction Decomposition into the simple
/
(x+1)/(x^2+1)
How do you (x+1)/(x^2+1) in partial fractions?
An expression to simplify:
Decompose fraction
The solution
You have entered
[src]
x + 1 ------ 2 x + 1
$$\frac{x + 1}{x^{2} + 1}$$
(x + 1)/(x^2 + 1)
Fraction decomposition
[src]
(1 + x)/(1 + x^2)
$$\frac{x + 1}{x^{2} + 1}$$
1 + x ------ 2 1 + x
Numerical answer
[src]
(1.0 + x)/(1.0 + x^2)
(1.0 + x)/(1.0 + x^2)