Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of x/(x^3+8)
Integral of x^(-6)
Integral of -x^-2
Integral of 1/(1+sqrt(x+1))
Identical expressions
x/(x+ one)*(2x+ one)
x divide by (x plus 1) multiply by (2x plus 1)
x divide by (x plus one) multiply by (2x plus one)
x/(x+1)(2x+1)
x/x+12x+1
x divide by (x+1)*(2x+1)
x/(x+1)*(2x+1)dx
Similar expressions
x/(x+1)(2x+1)
x/(x+1)*(2x-1)
(xdx)/((x+1)(2x+1))
x/(x-1)*(2x+1)

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

The solution

You have entered [src]

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

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

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

Detail solution

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

$x^{2} - x + \log{\left(x + 1 \right)}+ \mathrm{constant}$

The answer is:

$x^{2} - x + \log{\left(x + 1 \right)}+ \mathrm{constant}$

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

log(2)

$$\log{\left(2 \right)}$$

log(2)

$$\log{\left(2 \right)}$$

log(2)

Numerical answer [src]

0.693147180559945

0.693147180559945

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3