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Identical expressions
(one -y^ two)/(y+y^ three)
(1 minus y squared ) divide by (y plus y cubed )
(one minus y to the power of two) divide by (y plus y to the power of three)
(1-y2)/(y+y3)
1-y2/y+y3
(1-y²)/(y+y³)
(1-y to the power of 2)/(y+y to the power of 3)
1-y^2/y+y^3
(1-y^2) divide by (y+y^3)
Similar expressions
(1-y^2)/(y-y^3)
(1+y^2)/(y+y^3)

Solving integrals/ (1-y^2)/(y+y^3)

Integral of (1-y^2)/(y+y^3) dx

The solution

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  1          
  /          
 |           
 |       2   
 |  1 - y    
 |  ------ dy
 |       3   
 |  y + y    
 |           
/            
0

$$\int\limits_{0}^{1} \frac{1 - y^{2}}{y^{3} + y}\, dy$$

Integral((1 - y^2)/(y + y^3), (y, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{1 - y^{2}}{y^{3} + y} = - \frac{2 y}{y^{2} + 1} + \frac{1}{y}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{2 y}{y^{2} + 1}\right)\, dy = - 2 \int \frac{y}{y^{2} + 1}\, dy$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{y}{y^{2} + 1}\, dy = \frac{\int \frac{2 y}{y^{2} + 1}\, dy}{2}$
      
      Let $u = y^{2} + 1$ .
      
      Then let $du = 2 y dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(y^{2} + 1 \right)}$
      
      So, the result is: $\frac{\log{\left(y^{2} + 1 \right)}}{2}$
    So, the result is: $- \log{\left(y^{2} + 1 \right)}$
  1. The integral of $\frac{1}{y}$ is $\log{\left(y \right)}$ .
  The result is: $\log{\left(y \right)} - \log{\left(y^{2} + 1 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1 - y^{2}}{y^{3} + y} = - \frac{y^{2} - 1}{y^{3} + y}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{y^{2} - 1}{y^{3} + y}\right)\, dy = - \int \frac{y^{2} - 1}{y^{3} + y}\, dy$
  1. Rewrite the integrand:
    
    $\frac{y^{2} - 1}{y^{3} + y} = \frac{2 y}{y^{2} + 1} - \frac{1}{y}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 y}{y^{2} + 1}\, dy = 2 \int \frac{y}{y^{2} + 1}\, dy$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{y}{y^{2} + 1}\, dy = \frac{\int \frac{2 y}{y^{2} + 1}\, dy}{2}$
      
      Let $u = y^{2} + 1$ .
      
      Then let $du = 2 y dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(y^{2} + 1 \right)}$
      
      So, the result is: $\frac{\log{\left(y^{2} + 1 \right)}}{2}$
      
      So, the result is: $\log{\left(y^{2} + 1 \right)}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{y}\right)\, dy = - \int \frac{1}{y}\, dy$
      
      The integral of $\frac{1}{y}$ is $\log{\left(y \right)}$ .
      
      So, the result is: $- \log{\left(y \right)}$
    The result is: $- \log{\left(y \right)} + \log{\left(y^{2} + 1 \right)}$
  So, the result is: $\log{\left(y \right)} - \log{\left(y^{2} + 1 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1 - y^{2}}{y^{3} + y} = - \frac{y^{2}}{y^{3} + y} + \frac{1}{y^{3} + y}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{y^{2}}{y^{3} + y}\right)\, dy = - \int \frac{y^{2}}{y^{3} + y}\, dy$
    1. Rewrite the integrand:
      
      $\frac{y^{2}}{y^{3} + y} = \frac{y}{y^{2} + 1}$
    2. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{y}{y^{2} + 1}\, dy = \frac{\int \frac{2 y}{y^{2} + 1}\, dy}{2}$
      
      Let $u = y^{2} + 1$ .
      
      Then let $du = 2 y dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(y^{2} + 1 \right)}$
      
      So, the result is: $\frac{\log{\left(y^{2} + 1 \right)}}{2}$
    So, the result is: $- \frac{\log{\left(y^{2} + 1 \right)}}{2}$
  1. Rewrite the integrand:
    
    $\frac{1}{y^{3} + y} = - \frac{y}{y^{2} + 1} + \frac{1}{y}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{y}{y^{2} + 1}\right)\, dy = - \int \frac{y}{y^{2} + 1}\, dy$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{y}{y^{2} + 1}\, dy = \frac{\int \frac{2 y}{y^{2} + 1}\, dy}{2}$
      
      Let $u = y^{2} + 1$ .
      
      Then let $du = 2 y dy$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{1}{2 u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(y^{2} + 1 \right)}$
      
      So, the result is: $\frac{\log{\left(y^{2} + 1 \right)}}{2}$
      
      So, the result is: $- \frac{\log{\left(y^{2} + 1 \right)}}{2}$
    1. The integral of $\frac{1}{y}$ is $\log{\left(y \right)}$ .
    The result is: $\log{\left(y \right)} - \frac{\log{\left(y^{2} + 1 \right)}}{2}$
  The result is: $\log{\left(y \right)} - \log{\left(y^{2} + 1 \right)}$
Add the constant of integration:

$\log{\left(y \right)} - \log{\left(y^{2} + 1 \right)}+ \mathrm{constant}$

The answer is:

$\log{\left(y \right)} - \log{\left(y^{2} + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                    
 |                                     
 |      2                              
 | 1 - y              /     2\         
 | ------ dy = C - log\1 + y / + log(y)
 |      3                              
 | y + y                               
 |                                     
/

$$\int \frac{1 - y^{2}}{y^{3} + y}\, dy = C + \log{\left(y \right)} - \log{\left(y^{2} + 1 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

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$$\infty$$

oo

$$\infty$$

oo

Numerical answer [src]

43.3972989534329

43.3972989534329

The graph

Use the examples entering the upper and lower limits of integration.

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Identical expressions

Similar expressions

Integral of (1-y^2)/(y+y^3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3