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Integral of d{x}:
Integral of 1/(1+x^4)
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Integral of y=x
Derivative of:
1/(x+3/4)
Identical expressions
one /(x+ three / four)
1 divide by (x plus 3 divide by 4)
one divide by (x plus three divide by four)
1/x+3/4
1 divide by (x+3 divide by 4)
1/(x+3/4)dx
Similar expressions
1/(x-3/4)

Integral of 1/(x+3/4) dx

The solution

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  1           
  /           
 |            
 |     1      
 |  ------- dx
 |  x + 3/4   
 |            
/             
0

$$\int\limits_{0}^{1} \frac{1}{x + \frac{3}{4}}\, dx$$

Integral(1/(x + 3/4), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = x + \frac{3}{4}$ .
  
  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:
  
  $\log{\left(x + \frac{3}{4} \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1}{x + \frac{3}{4}} = \frac{4}{4 x + 3}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{4}{4 x + 3}\, dx = 4 \int \frac{1}{4 x + 3}\, dx$
  1. Let $u = 4 x + 3$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(4 x + 3 \right)}}{4}$
  So, the result is: $\log{\left(4 x + 3 \right)}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1}{x + \frac{3}{4}} = \frac{4}{4 x + 3}$
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{4}{4 x + 3}\, dx = 4 \int \frac{1}{4 x + 3}\, dx$
  1. Let $u = 4 x + 3$ .
    
    Then let $du = 4 dx$ and substitute $\frac{du}{4}$ :
    
    $\int \frac{1}{4 u}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{4}$
    Now substitute $u$ back in:
    
    $\frac{\log{\left(4 x + 3 \right)}}{4}$
  So, the result is: $\log{\left(4 x + 3 \right)}$
Now simplify:

$\log{\left(x + \frac{3}{4} \right)}$
Add the constant of integration:

$\log{\left(x + \frac{3}{4} \right)}+ \mathrm{constant}$

The answer is:

$\log{\left(x + \frac{3}{4} \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                             
 |                              
 |    1                         
 | ------- dx = C + log(x + 3/4)
 | x + 3/4                      
 |                              
/

$$\int \frac{1}{x + \frac{3}{4}}\, dx = C + \log{\left(x + \frac{3}{4} \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

-log(3) + log(7)

$$- \log{\left(3 \right)} + \log{\left(7 \right)}$$

-log(3) + log(7)

$$- \log{\left(3 \right)} + \log{\left(7 \right)}$$

-log(3) + log(7)

Numerical answer [src]

0.847297860387204

0.847297860387204

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

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Identical expressions

Similar expressions

Integral of 1/(x+3/4) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3