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Integral of d{x}:
Integral of ((2x^2+41x-91))/((x-1)(x+3)(x-4))
Integral of ctgx
Integral of x^2cosnx
Integral of sqrt((1+cot^(2)x))
Identical expressions
((two x^2+ four one x- ninety-one))/((x-1)(x+ three)(x-4))
((2x squared plus 41x minus 91)) divide by ((x minus 1)(x plus 3)(x minus 4))
((two x squared plus four one x minus ninety minus one)) divide by ((x minus 1)(x plus three)(x minus 4))
((2x2+41x-91))/((x-1)(x+3)(x-4))
2x2+41x-91/x-1x+3x-4
((2x²+41x-91))/((x-1)(x+3)(x-4))
((2x to the power of 2+41x-91))/((x-1)(x+3)(x-4))
2x^2+41x-91/x-1x+3x-4
((2x^2+41x-91)) divide by ((x-1)(x+3)(x-4))
((2x^2+41x-91))/((x-1)(x+3)(x-4))dx
Similar expressions
((2x^2+41x-91))/((x+1)(x+3)(x-4))
((2x^2+41x-91))/((x-1)(x-3)(x-4))
((2x^2-41x-91))/((x-1)(x+3)(x-4))
((2x^2+41x-91))/((x-1)(x+3)(x+4))
((2x^2+41x+91))/((x-1)(x+3)(x-4))

Solving integrals/ ((2x^2+41x-91))/((x-1)(x+3)(x-4))

Integral of ((2x^2+41x-91))/((x-1)(x+3)(x-4)) dx

The solution

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  1                           
  /                           
 |                            
 |         2                  
 |      2*x  + 41*x - 91      
 |  ----------------------- dx
 |  (x - 1)*(x + 3)*(x - 4)   
 |                            
/                             
0

$$\int\limits_{0}^{1} \frac{2 x^{2} + 41 x - 91}{\left(x + 3\right) \left(x - 1\right) \left(x - 4\right)}\, dx$$

Integral((2*x^2 + 41*x - 1*91)/(((x - 1*1)*(x + 3)*(x - 1*4))), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{2 x^{2} + 41 x - 91}{\left(x + 3\right) \left(x - 1\right) \left(x - 4\right)} = - \frac{7}{x + 3} + \frac{4}{x - 1} + \frac{5}{x - 4}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{7}{x + 3}\right)\, dx = - 7 \int \frac{1}{x + 3}\, dx$
    1. Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
    So, the result is: $- 7 \log{\left(x + 3 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4}{x - 1}\, dx = 4 \int \frac{1}{x - 1}\, dx$
    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: $4 \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{5}{x - 4}\, dx = 5 \int \frac{1}{x - 4}\, dx$
    1. Let $u = x - 4$ .
      
      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:
      
      $\log{\left(x - 4 \right)}$
    So, the result is: $5 \log{\left(x - 4 \right)}$
  The result is: $5 \log{\left(x - 4 \right)} + 4 \log{\left(x - 1 \right)} - 7 \log{\left(x + 3 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{2 x^{2} + 41 x - 91}{\left(x + 3\right) \left(x - 1\right) \left(x - 4\right)} = \frac{2 x^{2} + 41 x - 91}{x^{3} - 2 x^{2} - 11 x + 12}$
2. Rewrite the integrand:
  
  $\frac{2 x^{2} + 41 x - 91}{x^{3} - 2 x^{2} - 11 x + 12} = - \frac{7}{x + 3} + \frac{4}{x - 1} + \frac{5}{x - 4}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{7}{x + 3}\right)\, dx = - 7 \int \frac{1}{x + 3}\, dx$
    1. Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
    So, the result is: $- 7 \log{\left(x + 3 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{4}{x - 1}\, dx = 4 \int \frac{1}{x - 1}\, dx$
    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: $4 \log{\left(x - 1 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{5}{x - 4}\, dx = 5 \int \frac{1}{x - 4}\, dx$
    1. Let $u = x - 4$ .
      
      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:
      
      $\log{\left(x - 4 \right)}$
    So, the result is: $5 \log{\left(x - 4 \right)}$
  The result is: $5 \log{\left(x - 4 \right)} + 4 \log{\left(x - 1 \right)} - 7 \log{\left(x + 3 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{2 x^{2} + 41 x - 91}{\left(x + 3\right) \left(x - 1\right) \left(x - 4\right)} = \frac{2 x^{2}}{x^{3} - 2 x^{2} - 11 x + 12} + \frac{41 x}{x^{3} - 2 x^{2} - 11 x + 12} - \frac{91}{x^{3} - 2 x^{2} - 11 x + 12}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2 x^{2}}{x^{3} - 2 x^{2} - 11 x + 12}\, dx = 2 \int \frac{x^{2}}{x^{3} - 2 x^{2} - 11 x + 12}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x^{2}}{x^{3} - 2 x^{2} - 11 x + 12} = \frac{9}{28 \left(x + 3\right)} - \frac{1}{12 \left(x - 1\right)} + \frac{16}{21 \left(x - 4\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{9}{28 \left(x + 3\right)}\, dx = \frac{9 \int \frac{1}{x + 3}\, dx}{28}$
      
      Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
      
      So, the result is: $\frac{9 \log{\left(x + 3 \right)}}{28}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{12 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{12}$
      
      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: $- \frac{\log{\left(x - 1 \right)}}{12}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{16}{21 \left(x - 4\right)}\, dx = \frac{16 \int \frac{1}{x - 4}\, dx}{21}$
      
      Let $u = x - 4$ .
      
      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:
      
      $\log{\left(x - 4 \right)}$
      
      So, the result is: $\frac{16 \log{\left(x - 4 \right)}}{21}$
      
      The result is: $\frac{16 \log{\left(x - 4 \right)}}{21} - \frac{\log{\left(x - 1 \right)}}{12} + \frac{9 \log{\left(x + 3 \right)}}{28}$
    So, the result is: $\frac{32 \log{\left(x - 4 \right)}}{21} - \frac{\log{\left(x - 1 \right)}}{6} + \frac{9 \log{\left(x + 3 \right)}}{14}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{41 x}{x^{3} - 2 x^{2} - 11 x + 12}\, dx = 41 \int \frac{x}{x^{3} - 2 x^{2} - 11 x + 12}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x}{x^{3} - 2 x^{2} - 11 x + 12} = - \frac{3}{28 \left(x + 3\right)} - \frac{1}{12 \left(x - 1\right)} + \frac{4}{21 \left(x - 4\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3}{28 \left(x + 3\right)}\right)\, dx = - \frac{3 \int \frac{1}{x + 3}\, dx}{28}$
      
      Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
      
      So, the result is: $- \frac{3 \log{\left(x + 3 \right)}}{28}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{12 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{12}$
      
      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: $- \frac{\log{\left(x - 1 \right)}}{12}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{4}{21 \left(x - 4\right)}\, dx = \frac{4 \int \frac{1}{x - 4}\, dx}{21}$
      
      Let $u = x - 4$ .
      
      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:
      
      $\log{\left(x - 4 \right)}$
      
      So, the result is: $\frac{4 \log{\left(x - 4 \right)}}{21}$
      
      The result is: $\frac{4 \log{\left(x - 4 \right)}}{21} - \frac{\log{\left(x - 1 \right)}}{12} - \frac{3 \log{\left(x + 3 \right)}}{28}$
    So, the result is: $\frac{164 \log{\left(x - 4 \right)}}{21} - \frac{41 \log{\left(x - 1 \right)}}{12} - \frac{123 \log{\left(x + 3 \right)}}{28}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{91}{x^{3} - 2 x^{2} - 11 x + 12}\right)\, dx = - 91 \int \frac{1}{x^{3} - 2 x^{2} - 11 x + 12}\, dx$
    1. Rewrite the integrand:
      
      $\frac{1}{x^{3} - 2 x^{2} - 11 x + 12} = \frac{1}{28 \left(x + 3\right)} - \frac{1}{12 \left(x - 1\right)} + \frac{1}{21 \left(x - 4\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{28 \left(x + 3\right)}\, dx = \frac{\int \frac{1}{x + 3}\, dx}{28}$
      
      Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
      
      So, the result is: $\frac{\log{\left(x + 3 \right)}}{28}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{12 \left(x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{x - 1}\, dx}{12}$
      
      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: $- \frac{\log{\left(x - 1 \right)}}{12}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{21 \left(x - 4\right)}\, dx = \frac{\int \frac{1}{x - 4}\, dx}{21}$
      
      Let $u = x - 4$ .
      
      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:
      
      $\log{\left(x - 4 \right)}$
      
      So, the result is: $\frac{\log{\left(x - 4 \right)}}{21}$
      
      The result is: $\frac{\log{\left(x - 4 \right)}}{21} - \frac{\log{\left(x - 1 \right)}}{12} + \frac{\log{\left(x + 3 \right)}}{28}$
    So, the result is: $- \frac{13 \log{\left(x - 4 \right)}}{3} + \frac{91 \log{\left(x - 1 \right)}}{12} - \frac{13 \log{\left(x + 3 \right)}}{4}$
  The result is: $5 \log{\left(x - 4 \right)} + 4 \log{\left(x - 1 \right)} - 7 \log{\left(x + 3 \right)}$
Add the constant of integration:

$5 \log{\left(x - 4 \right)} + 4 \log{\left(x - 1 \right)} - 7 \log{\left(x + 3 \right)}+ \mathrm{constant}$

The answer is:

$5 \log{\left(x - 4 \right)} + 4 \log{\left(x - 1 \right)} - 7 \log{\left(x + 3 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                             
 |                                                                              
 |        2                                                                     
 |     2*x  + 41*x - 91                                                         
 | ----------------------- dx = C - 7*log(3 + x) + 4*log(-1 + x) + 5*log(-4 + x)
 | (x - 1)*(x + 3)*(x - 4)                                                      
 |                                                                              
/

$$-7\,\log \left(x+3\right)+4\,\log \left(x-1\right)+5\,\log \left(x- 4\right)$$

Simplify

The answer [src]

-oo - 9*pi*I

$${\it \%a}$$

-oo - 9*pi*I

$$-\infty - 9 i \pi$$

Simplify

Numerical answer [src]

-179.816012014299

-179.816012014299

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

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Identical expressions

Similar expressions

Integral of ((2x^2+41x-91))/((x-1)(x+3)(x-4)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3