Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of e^(-x)/sqrt(x)
Integral of 8/x^2
Integral of (1+x^2)^0.5
Integral of 2*t
Identical expressions
one /(2x+ one)log(2x+ one)
1 divide by (2x plus 1) logarithm of (2x plus 1)
one divide by (2x plus one) logarithm of (2x plus one)
1/2x+1log2x+1
1 divide by (2x+1)log(2x+1)
1/(2x+1)log(2x+1)dx
Similar expressions
1/(2x+1)log(2x-1)
1/(2x-1)log(2x+1)

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

The solution

You have entered [src]

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

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

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

Detail solution

Let $u = 2 x + 1$ .

Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :

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

$\frac{\log{\left(2 x + 1 \right)}^{2}}{4}$
Now simplify:

$\frac{\log{\left(2 x + 1 \right)}^{2}}{4}$
Add the constant of integration:

$\frac{\log{\left(2 x + 1 \right)}^{2}}{4}+ \mathrm{constant}$

The answer is:

$\frac{\log{\left(2 x + 1 \right)}^{2}}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                   
 |                          2         
 | log(2*x + 1)          log (2*x + 1)
 | ------------ dx = C + -------------
 |   2*x + 1                   4      
 |                                    
/

$$\int \frac{\log{\left(2 x + 1 \right)}}{2 x + 1}\, dx = C + \frac{\log{\left(2 x + 1 \right)}^{2}}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

   2   
log (3)
-------
   4

$$\frac{\log{\left(3 \right)}^{2}}{4}$$

   2   
log (3)
-------
   4

$$\frac{\log{\left(3 \right)}^{2}}{4}$$

log(3)^2/4

Numerical answer [src]

0.301737240203146

0.301737240203146

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2