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log(one +x)/sqrt(one +x)
logarithm of (1 plus x) divide by square root of (1 plus x)
logarithm of (one plus x) divide by square root of (one plus x)
log(1+x)/√(1+x)
log1+x/sqrt1+x
log(1+x) divide by sqrt(1+x)
log(1+x)/sqrt(1+x)dx
Similar expressions
log(1+x)/sqrt(1-x)
log(1-x)/sqrt(1+x)

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

The solution

You have entered [src]

  3              
  /              
 |               
 |  log(1 + x)   
 |  ---------- dx
 |    _______    
 |  \/ 1 + x     
 |               
/                
0

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

Integral(log(1 + x)/sqrt(1 + x), (x, 0, 3))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \log{\left(x + 1 \right)}$ .
  
  Then let $du = \frac{dx}{x + 1}$ and substitute $du$ :
  
  $\int u e^{\frac{u}{2}}\, du$
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(u \right)} = u$ and let $\operatorname{dv}{\left(u \right)} = e^{\frac{u}{2}}$ .
    
    Then $\operatorname{du}{\left(u \right)} = 1$ .
    
    To find $v{\left(u \right)}$ :
    1. Let $u = \frac{u}{2}$ .
      
      Then let $du = \frac{du}{2}$ and substitute $2 du$ :
      
      $\int 2 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{u}{2}}$
    Now evaluate the sub-integral.
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 e^{\frac{u}{2}}\, du = 2 \int e^{\frac{u}{2}}\, du$
    1. Let $u = \frac{u}{2}$ .
      
      Then let $du = \frac{du}{2}$ and substitute $2 du$ :
      
      $\int 2 e^{u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\text{False}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $2 e^{u}$
      
      Now substitute $u$ back in:
      
      $2 e^{\frac{u}{2}}$
    So, the result is: $4 e^{\frac{u}{2}}$
  Now substitute $u$ back in:
  
  $2 \sqrt{x + 1} \log{\left(x + 1 \right)} - 4 \sqrt{x + 1}$
Method #2
1. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(x \right)} = \log{\left(x + 1 \right)}$ and let $\operatorname{dv}{\left(x \right)} = \frac{1}{\sqrt{x + 1}}$ .
  
  Then $\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}$ .
  
  To find $v{\left(x \right)}$ :
  1. Let $u = x + 1$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{\sqrt{u}}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}$
    Now substitute $u$ back in:
    
    $2 \sqrt{x + 1}$
  Now evaluate the sub-integral.
2. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{2}{\sqrt{x + 1}}\, dx = 2 \int \frac{1}{\sqrt{x + 1}}\, dx$
  1. Let $u = x + 1$ .
    
    Then let $du = dx$ and substitute $du$ :
    
    $\int \frac{1}{\sqrt{u}}\, du$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}$
    Now substitute $u$ back in:
    
    $2 \sqrt{x + 1}$
  So, the result is: $4 \sqrt{x + 1}$
Now simplify:

$2 \sqrt{x + 1} \left(\log{\left(x + 1 \right)} - 2\right)$
Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

-4 + 4*log(4)

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

-4 + 4*log(4)

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

-4 + 4*log(4)

Numerical answer [src]

1.54517744447956

1.54517744447956

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

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Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2