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Identical expressions
x^ two +4x+ thirteen
x squared plus 4x plus 13
x to the power of two plus 4x plus thirteen
x2+4x+13
x²+4x+13
x to the power of 2+4x+13
x^2+4x+13dx
Similar expressions
x^2+4x-13
x^2-4x+13

Integral of x^2+4x+13 dx

The solution

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 oo                   
  /                   
 |                    
 |  / 2           \   
 |  \x  + 4*x + 13/ dx
 |                    
/                     
1

$$\int\limits_{1}^{\infty} \left(\left(x^{2} + 4 x\right) + 13\right)\, dx$$

Integral(x^2 + 4*x + 13, (x, 1, oo))

Detail solution

Integrate term-by-term:
1. Integrate term-by-term:
  1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
    
    $\int x^{2}\, dx = \frac{x^{3}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 4 x\, dx = 4 \int x\, dx$
    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: $2 x^{2}$
  The result is: $\frac{x^{3}}{3} + 2 x^{2}$
1. The integral of a constant is the constant times the variable of integration:
  
  $\int 13\, dx = 13 x$
The result is: $\frac{x^{3}}{3} + 2 x^{2} + 13 x$
Now simplify:

$\frac{x \left(x^{2} + 6 x + 39\right)}{3}$
Add the constant of integration:

$\frac{x \left(x^{2} + 6 x + 39\right)}{3}+ \mathrm{constant}$

The answer is:

$\frac{x \left(x^{2} + 6 x + 39\right)}{3}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                         
 |                                         3
 | / 2           \             2          x 
 | \x  + 4*x + 13/ dx = C + 2*x  + 13*x + --
 |                                        3 
/

$$\int \left(\left(x^{2} + 4 x\right) + 13\right)\, dx = C + \frac{x^{3}}{3} + 2 x^{2} + 13 x$$

Simplify

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The answer [src]

oo

$$\infty$$

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$$\infty$$

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

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Integral of x^2+4x+13 dx

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The solution