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Integral of d{x}:
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Identical expressions
x^3dx/(x^ two -3x+ two)
x cubed dx divide by (x squared minus 3x plus 2)
x cubed dx divide by (x to the power of two minus 3x plus two)
x3dx/(x2-3x+2)
x3dx/x2-3x+2
x³dx/(x²-3x+2)
x to the power of 3dx/(x to the power of 2-3x+2)
x^3dx/x^2-3x+2
x^3dx divide by (x^2-3x+2)
Similar expressions
x^3dx/(x^2-3x-2)
x^3dx/(x^2+3x+2)
What you mean?
x^3*dx/(x^2 - 3*x + 2)
x^3*dx/(x^2 - 3*x) + 2
x^3*dx*x/(x^2 - 1*3) + 2
x^3*dx/(x^2) - 3*x + 2
x^3*dx*2/x - 3*x + 2
x^3*dx/(x^2) - 3*x + 2
x^3*dx/(x^2) - 3*x + 2

You entered:

x^3dx/(x^2-3x+2)

What you mean?

x^3*dx/(x^2 - 3*x + 2)

x^3*dx/(x^2 - 3*x) + 2

x^3*dx*x/(x^2 - 1*3) + 2

x^3*dx/(x^2) - 3*x + 2

x^3*dx*2/x - 3*x + 2

x^3*dx/(x^2) - 3*x + 2

x^3*dx/(x^2) - 3*x + 2

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

The solution

You have entered [src]

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

$$\int\limits_{0}^{1} x^{3} \cdot 1 \cdot \frac{1}{x^{2} - 3 x + 2}\, dx$$

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

Detail solution

Rewrite the integrand:

$x^{3} \cdot 1 \cdot \frac{1}{x^{2} - 3 x + 2} = x + 3 - \frac{1}{x - 1} + \frac{8}{x - 2}$
Integrate term-by-term:
1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
  
  $\int x\, dx = \frac{x^{2}}{2}$
1. The integral of a constant is the constant times the variable of integration:
  
  $\int 3\, dx = 3 x$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \left(- \frac{1}{x - 1}\right)\, dx = - \int \frac{1}{x - 1}\, dx$
  1. Let $u = x - 1$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. 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: $- \log{\left(x - 1 \right)}$
1. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{8}{x - 2}\, dx = 8 \int \frac{1}{x - 2}\, dx$
  1. Let $u = x - 2$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{u}\, du$
    1. The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
    Now substitute $u$ back in:
    
    $\log{\left(x - 2 \right)}$
  So, the result is: $8 \log{\left(x - 2 \right)}$
The result is: $\frac{x^{2}}{2} + 3 x + 8 \log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}$
Add the constant of integration:

$\frac{x^{2}}{2} + 3 x + 8 \log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer is:

$\frac{x^{2}}{2} + 3 x + 8 \log{\left(x - 2 \right)} - \log{\left(x - 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                 
 |                             2                                    
 |  3        1                x                                     
 | x *1*------------ dx = C + -- - log(-1 + x) + 3*x + 8*log(-2 + x)
 |       2                    2                                     
 |      x  - 3*x + 2                                                
 |                                                                  
/

$${{x^2+6\,x}\over{2}}-\log \left(x-1\right)+8\,\log \left(x-2\right)$$

Simplify

The answer [src]

oo - 7*pi*I

$${\it \%a}$$

oo - 7*pi*I

$$\infty - 7 i \pi$$

Упростить

Numerical answer [src]

42.0441784776431

42.0441784776431

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

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What you mean?

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

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The solution