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Identical expressions
sin(log(x))/x+ four hundred and four
sinus of ( logarithm of (x)) divide by x plus 404
sinus of ( logarithm of (x)) divide by x plus four hundred and four
sinlogx/x+404
sin(log(x)) divide by x+404
sin(log(x))/x+404dx
Similar expressions
sin(log(x))/x-404

Integral of sin(log(x))/x+404 dx

The solution

You have entered [src]

  1                       
  /                       
 |                        
 |  /sin(log(x))      \   
 |  |----------- + 404| dx
 |  \     x           /   
 |                        
/                         
0

$$\int\limits_{0}^{1} \left(404 + \frac{\sin{\left(\log{\left(x \right)} \right)}}{x}\right)\, dx$$

Integral(sin(log(x))/x + 404, (x, 0, 1))

Detail solution

Integrate term-by-term:
1. The integral of a constant is the constant times the variable of integration:
  
  $\int 404\, dx = 404 x$
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int \sin{\left(u \right)}\, du$
    1. The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
    Now substitute $u$ back in:
    
    $- \cos{\left(\log{\left(x \right)} \right)}$
  Method #2
  1. Let $u = \frac{1}{x}$ .
    
    Then let $du = - \frac{dx}{x^{2}}$ and substitute $- du$ :
    
    $\int \left(- \frac{\sin{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\right)\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{\sin{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\, du = - \int \frac{\sin{\left(\log{\left(\frac{1}{u} \right)} \right)}}{u}\, du$
      
      Let $u = \log{\left(\frac{1}{u} \right)}$ .
      
      Then let $du = - \frac{du}{u}$ and substitute $- du$ :
      
      $\int \left(- \sin{\left(u \right)}\right)\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sin{\left(u \right)}\, du = - \int \sin{\left(u \right)}\, du$
      
      The integral of sine is negative cosine:
      
      $\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}$
      
      So, the result is: $\cos{\left(u \right)}$
      
      Now substitute $u$ back in:
      
      $\cos{\left(\log{\left(\frac{1}{u} \right)} \right)}$
      
      So, the result is: $- \cos{\left(\log{\left(\frac{1}{u} \right)} \right)}$
    Now substitute $u$ back in:
    
    $- \cos{\left(\log{\left(x \right)} \right)}$
The result is: $404 x - \cos{\left(\log{\left(x \right)} \right)}$
Add the constant of integration:

$404 x - \cos{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}$

The answer is:

$404 x - \cos{\left(\log{\left(x \right)} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                
 |                                                 
 | /sin(log(x))      \                             
 | |----------- + 404| dx = C - cos(log(x)) + 404*x
 | \     x           /                             
 |                                                 
/

$$\int \left(404 + \frac{\sin{\left(\log{\left(x \right)} \right)}}{x}\right)\, dx = C + 404 x - \cos{\left(\log{\left(x \right)} \right)}$$

Simplify

The answer [src]

<402, 404>

$$\left\langle 402, 404\right\rangle$$

<402, 404>

$$\left\langle 402, 404\right\rangle$$

AccumBounds(402, 404)

Numerical answer [src]

404.014150063156

404.014150063156

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

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Integral of sin(log(x))/x+404 dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2