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Identical expressions
x/(one +sqrt(2x- one))
x divide by (1 plus square root of (2x minus 1))
x divide by (one plus square root of (2x minus one))
x/(1+√(2x-1))
x/1+sqrt2x-1
x divide by (1+sqrt(2x-1))
x/(1+sqrt(2x-1))dx
Similar expressions
x/(1+sqrt(2x+1))
(dx)/(1+sqrt(2x-1))
x/(1-sqrt(2x-1))

Integral of x/(1+sqrt(2x-1)) dx

The solution

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  1                   
  /                   
 |                    
 |         x          
 |  --------------- dx
 |        _________   
 |  1 + \/ 2*x - 1    
 |                    
/                     
0

$$\int\limits_{0}^{1} \frac{x}{\sqrt{2 x - 1} + 1}\, dx$$

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

Detail solution

Let $u = \sqrt{2 x - 1}$ .

Then let $du = \frac{dx}{\sqrt{2 x - 1}}$ and substitute $du$ :

$\int \frac{u \left(\frac{u^{2}}{2} + \frac{1}{2}\right)}{u + 1}\, du$
1. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\frac{u \left(\frac{u^{2}}{2} + \frac{1}{2}\right)}{u + 1} = \frac{u^{2}}{2} - \frac{u}{2} + 1 - \frac{1}{u + 1}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $\frac{u^{3}}{6}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u}{2}\right)\, du = - \frac{\int u\, du}{2}$
      
      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: $- \frac{u^{2}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
      
      So, the result is: $- \log{\left(u + 1 \right)}$
    The result is: $\frac{u^{3}}{6} - \frac{u^{2}}{4} + u - \log{\left(u + 1 \right)}$
  Method #2
  1. Rewrite the integrand:
    
    $\frac{u \left(\frac{u^{2}}{2} + \frac{1}{2}\right)}{u + 1} = \frac{u^{3} + u}{2 u + 2}$
  2. Rewrite the integrand:
    
    $\frac{u^{3} + u}{2 u + 2} = \frac{u^{2}}{2} - \frac{u}{2} + 1 - \frac{1}{u + 1}$
  3. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      So, the result is: $\frac{u^{3}}{6}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{u}{2}\right)\, du = - \frac{\int u\, du}{2}$
      
      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: $- \frac{u^{2}}{4}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
      
      So, the result is: $- \log{\left(u + 1 \right)}$
    The result is: $\frac{u^{3}}{6} - \frac{u^{2}}{4} + u - \log{\left(u + 1 \right)}$
  Method #3
  1. Rewrite the integrand:
    
    $\frac{u \left(\frac{u^{2}}{2} + \frac{1}{2}\right)}{u + 1} = \frac{u^{3}}{2 \left(u + 1\right)} + \frac{u}{2 \left(u + 1\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{3}}{2 \left(u + 1\right)}\, du = \frac{\int \frac{u^{3}}{u + 1}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u^{3}}{u + 1} = u^{2} - u + 1 - \frac{1}{u + 1}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{2}\, du = \frac{u^{3}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u\right)\, du = - \int u\, du$
      
      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: $- \frac{u^{2}}{2}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
      
      So, the result is: $- \log{\left(u + 1 \right)}$
      
      The result is: $\frac{u^{3}}{3} - \frac{u^{2}}{2} + u - \log{\left(u + 1 \right)}$
      
      So, the result is: $\frac{u^{3}}{6} - \frac{u^{2}}{4} + \frac{u}{2} - \frac{\log{\left(u + 1 \right)}}{2}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u}{2 \left(u + 1\right)}\, du = \frac{\int \frac{u}{u + 1}\, du}{2}$
      
      Rewrite the integrand:
      
      $\frac{u}{u + 1} = 1 - \frac{1}{u + 1}$
      
      Integrate term-by-term:
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du$
      
      Let $u = u + 1$ .
      
      Then let $du = du$ and substitute $du$ :
      
      $\int \frac{1}{u}\, du$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      Now substitute $u$ back in:
      
      $\log{\left(u + 1 \right)}$
      
      So, the result is: $- \log{\left(u + 1 \right)}$
      
      The result is: $u - \log{\left(u + 1 \right)}$
      
      So, the result is: $\frac{u}{2} - \frac{\log{\left(u + 1 \right)}}{2}$
    The result is: $\frac{u^{3}}{6} - \frac{u^{2}}{4} + u - \log{\left(u + 1 \right)}$
Now substitute $u$ back in:

$- \frac{x}{2} + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{6} + \sqrt{2 x - 1} - \log{\left(\sqrt{2 x - 1} + 1 \right)} + \frac{1}{4}$
Now simplify:

$- \frac{x}{2} + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{6} + \sqrt{2 x - 1} - \log{\left(\sqrt{2 x - 1} + 1 \right)} + \frac{1}{4}$
Add the constant of integration:

$- \frac{x}{2} + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{6} + \sqrt{2 x - 1} - \log{\left(\sqrt{2 x - 1} + 1 \right)} + \frac{1}{4}+ \mathrm{constant}$

The answer is:

$- \frac{x}{2} + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{6} + \sqrt{2 x - 1} - \log{\left(\sqrt{2 x - 1} + 1 \right)} + \frac{1}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                  
 |                                                                                3/2
 |        x             1         _________      /      _________\   x   (2*x - 1)   
 | --------------- dx = - + C + \/ 2*x - 1  - log\1 + \/ 2*x - 1 / - - + ------------
 |       _________      4                                            2        6      
 | 1 + \/ 2*x - 1                                                                    
 |                                                                                   
/

$$\int \frac{x}{\sqrt{2 x - 1} + 1}\, dx = C - \frac{x}{2} + \frac{\left(2 x - 1\right)^{\frac{3}{2}}}{6} + \sqrt{2 x - 1} - \log{\left(\sqrt{2 x - 1} + 1 \right)} + \frac{1}{4}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

2            5*I   pi*I      /  ___\
- - log(2) - --- + ---- + log\\/ 2 /
3             6     4

$$- \log{\left(2 \right)} + \log{\left(\sqrt{2} \right)} + \frac{2}{3} - \frac{5 i}{6} + \frac{i \pi}{4}$$

2            5*I   pi*I      /  ___\
- - log(2) - --- + ---- + log\\/ 2 /
3             6     4

$$- \log{\left(2 \right)} + \log{\left(\sqrt{2} \right)} + \frac{2}{3} - \frac{5 i}{6} + \frac{i \pi}{4}$$

2/3 - log(2) - 5*i/6 + pi*i/4 + log(sqrt(2))

Numerical answer [src]

(0.320294110548451 - 0.0477365409496419j)

(0.320294110548451 - 0.0477365409496419j)

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

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How to use it?

Identical expressions

Similar expressions

Integral of x/(1+sqrt(2x-1)) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3