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Integral of (1+x^4)^(1/2)
Graphing y =:
(x+3)/(x-4)
Identical expressions
(x+ three)/(x- four)
(x plus 3) divide by (x minus 4)
(x plus three) divide by (x minus four)
x+3/x-4
(x+3) divide by (x-4)
(x+3)/(x-4)dx
Similar expressions
(x-3)/(x-4)
(x+3)/(x+4)

Solving integrals/ (x+3)/(x-4)

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

The solution

You have entered [src]

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

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

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{x + 3}{x - 4} = 1 + \frac{7}{x - 4}$
2. Integrate term-by-term:
  1. The integral of a constant is the constant times the variable of integration:
    
    $\int 1\, dx = x$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{7}{x - 4}\, dx = 7 \int \frac{1}{x - 4}\, dx$
    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: $7 \log{\left(x - 4 \right)}$
  The result is: $x + 7 \log{\left(x - 4 \right)}$
Method #2
1. Rewrite the integrand:
  
  $\frac{x + 3}{x - 4} = \frac{x}{x - 4} + \frac{3}{x - 4}$
2. Integrate term-by-term:
  1. Rewrite the integrand:
    
    $\frac{x}{x - 4} = 1 + \frac{4}{x - 4}$
  2. Integrate term-by-term:
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, dx = x$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{4}{x - 4}\, dx = 4 \int \frac{1}{x - 4}\, dx$
      
      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: $4 \log{\left(x - 4 \right)}$
    The result is: $x + 4 \log{\left(x - 4 \right)}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3}{x - 4}\, dx = 3 \int \frac{1}{x - 4}\, dx$
    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: $3 \log{\left(x - 4 \right)}$
  The result is: $x + 3 \log{\left(x - 4 \right)} + 4 \log{\left(x - 4 \right)}$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

  /                                
 |                                 
 | x + 3                           
 | ----- dx = C + x + 7*log(-4 + x)
 | x - 4                           
 |                                 
/

$$\int \frac{x + 3}{x - 4}\, dx = C + x + 7 \log{\left(x - 4 \right)}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

1 - 7*log(4) + 7*log(3)

$$- 7 \log{\left(4 \right)} + 1 + 7 \log{\left(3 \right)}$$

1 - 7*log(4) + 7*log(3)

$$- 7 \log{\left(4 \right)} + 1 + 7 \log{\left(3 \right)}$$

1 - 7*log(4) + 7*log(3)

Numerical answer [src]

-1.01377450716247

-1.01377450716247

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2