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(two x^2+2x- three)/(x+ one)
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(2x^2+2x-3)/(x+1)dx
Similar expressions
(2x^2-2x-3)/(x+1)
(2x^2+2x+3)/(x+1)
(2x^2+2x-3)/(x-1)

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

The solution

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  1                  
  /                  
 |                   
 |     2             
 |  2*x  + 2*x - 3   
 |  -------------- dx
 |      x + 1        
 |                   
/                    
0

$$\int\limits_{0}^{1} \frac{\left(2 x^{2} + 2 x\right) - 3}{x + 1}\, dx$$

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

Detail solution

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

$x^{2} - 3 \log{\left(x + 1 \right)}+ \mathrm{constant}$

The answer is:

$x^{2} - 3 \log{\left(x + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                         
 |                                          
 |    2                                     
 | 2*x  + 2*x - 3           2               
 | -------------- dx = C + x  - 3*log(1 + x)
 |     x + 1                                
 |                                          
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

1 - 3*log(2)

$$1 - 3 \log{\left(2 \right)}$$

1 - 3*log(2)

$$1 - 3 \log{\left(2 \right)}$$

1 - 3*log(2)

Numerical answer [src]

-1.07944154167984

-1.07944154167984

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2