Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of u^(-2)
Integral of e^(2*x)/2
Integral of 1/sqrt(1-x)
Integral of sin(x²)
Identical expressions
one /(x*(x+ two)*(x+ three))
1 divide by (x multiply by (x plus 2) multiply by (x plus 3))
one divide by (x multiply by (x plus two) multiply by (x plus three))
1/(x(x+2)(x+3))
1/xx+2x+3
1 divide by (x*(x+2)*(x+3))
1/(x*(x+2)*(x+3))dx
Similar expressions
1/(x*(x-2)*(x+3))
1/(x*(x+2)*(x-3))

Integral of 1/(x(x+2)(x+3)) dx

The solution

You have entered [src]

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

\int\limits_{0}^{1} \frac{1}{x \left(x + 2\right) \left(x + 3\right)}\, dx

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

Detail solution

There are multiple ways to do this integral.
Method #1
1. Rewrite the integrand:
  
  $\frac{1}{x \left(x + 2\right) \left(x + 3\right)} = \frac{1}{3 \left(x + 3\right)} - \frac{1}{2 \left(x + 2\right)} + \frac{1}{6 x}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{3 \left(x + 3\right)}\, dx = \frac{\int \frac{1}{x + 3}\, dx}{3}$
    1. Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
    So, the result is: $\frac{\log{\left(x + 3 \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 2\right)}\right)\, dx = - \frac{\int \frac{1}{x + 2}\, dx}{2}$
    1. Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $- \frac{\log{\left(x + 2 \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{6 x}\, dx = \frac{\int \frac{1}{x}\, dx}{6}$
    1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
    So, the result is: $\frac{\log{\left(x \right)}}{6}$
  The result is: $\frac{\log{\left(x \right)}}{6} - \frac{\log{\left(x + 2 \right)}}{2} + \frac{\log{\left(x + 3 \right)}}{3}$
Method #2
1. Rewrite the integrand:
  
  $\frac{1}{x \left(x + 2\right) \left(x + 3\right)} = \frac{1}{x^{3} + 5 x^{2} + 6 x}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{3} + 5 x^{2} + 6 x} = \frac{1}{3 \left(x + 3\right)} - \frac{1}{2 \left(x + 2\right)} + \frac{1}{6 x}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{3 \left(x + 3\right)}\, dx = \frac{\int \frac{1}{x + 3}\, dx}{3}$
    1. Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
    So, the result is: $\frac{\log{\left(x + 3 \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 2\right)}\right)\, dx = - \frac{\int \frac{1}{x + 2}\, dx}{2}$
    1. Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $- \frac{\log{\left(x + 2 \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{6 x}\, dx = \frac{\int \frac{1}{x}\, dx}{6}$
    1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
    So, the result is: $\frac{\log{\left(x \right)}}{6}$
  The result is: $\frac{\log{\left(x \right)}}{6} - \frac{\log{\left(x + 2 \right)}}{2} + \frac{\log{\left(x + 3 \right)}}{3}$
Method #3
1. Rewrite the integrand:
  
  $\frac{1}{x \left(x + 2\right) \left(x + 3\right)} = \frac{1}{x^{3} + 5 x^{2} + 6 x}$
2. Rewrite the integrand:
  
  $\frac{1}{x^{3} + 5 x^{2} + 6 x} = \frac{1}{3 \left(x + 3\right)} - \frac{1}{2 \left(x + 2\right)} + \frac{1}{6 x}$
3. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{3 \left(x + 3\right)}\, dx = \frac{\int \frac{1}{x + 3}\, dx}{3}$
    1. Let $u = x + 3$ .
      
      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:
      
      $\log{\left(x + 3 \right)}$
    So, the result is: $\frac{\log{\left(x + 3 \right)}}{3}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{1}{2 \left(x + 2\right)}\right)\, dx = - \frac{\int \frac{1}{x + 2}\, dx}{2}$
    1. Let $u = x + 2$ .
      
      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:
      
      $\log{\left(x + 2 \right)}$
    So, the result is: $- \frac{\log{\left(x + 2 \right)}}{2}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{1}{6 x}\, dx = \frac{\int \frac{1}{x}\, dx}{6}$
    1. The integral of $\frac{1}{x}$ is $\log{\left(x \right)}$ .
    So, the result is: $\frac{\log{\left(x \right)}}{6}$
  The result is: $\frac{\log{\left(x \right)}}{6} - \frac{\log{\left(x + 2 \right)}}{2} + \frac{\log{\left(x + 3 \right)}}{3}$
Add the constant of integration:

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

The answer is:

\frac{\log{\left(x \right)}}{6} - \frac{\log{\left(x + 2 \right)}}{2} + \frac{\log{\left(x + 3 \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src]

  /                                                           
 |                                                            
 |         1                  log(2 + x)   log(3 + x)   log(x)
 | ----------------- dx = C - ---------- + ---------- + ------
 | x*(x + 2)*(x + 3)              2            3          6   
 |                                                            
/

\int \frac{1}{x \left(x + 2\right) \left(x + 3\right)}\, dx = C + \frac{\log{\left(x \right)}}{6} - \frac{\log{\left(x + 2 \right)}}{2} + \frac{\log{\left(x + 3 \right)}}{3}

Simplify

The graph

Plot the graph f(x)

The answer [src]

oo

\infty

oo

\infty

oo

Numerical answer [src]

7.24156915909533

7.24156915909533

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of 1/(x(x+2)(x+3)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of 1/(x*(x+2)*(x+3)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3

Integral of 1/(x(x+2)(x+3)) dx