Mister Exam

Lang:

Other calculators

How to use it?
Integral of d{x}:
Integral of -e^x
Integral of (1+x^4)^(1/2)
Integral of (x^3+2)/(x^3-4x)
Integral of sqr(x)
Identical expressions
x/(sqrt(2x+ one)+ one)
x divide by ( square root of (2x plus 1) plus 1)
x divide by ( square root of (2x plus one) plus one)
x/(√(2x+1)+1)
x/sqrt2x+1+1
x divide by (sqrt(2x+1)+1)
x/(sqrt(2x+1)+1)dx
Similar expressions
x/((sqrt(2x+1))+1)
(x)/(sqrt(2x+1)+1)
x/(sqrt(2x+1)-1)
x/(sqrt(2x-1)+1)

Solving integrals/ x/(sqrt(2x+1)+1)

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

The solution

You have entered [src]

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

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

Integral(x/(sqrt(2*x + 1) + 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:
    
    $\frac{u \left(\frac{u^{2}}{2} - \frac{1}{2}\right)}{u + 1} = \frac{u^{2}}{2} - \frac{u}{2}$
  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: $- \frac{u^{2}}{4}$
    The result is: $\frac{u^{3}}{6} - \frac{u^{2}}{4}$
  Method #2
  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:
      
      $\int \left(- \frac{u}{2 \left(u + 1\right)}\right)\, du = - \frac{\int \frac{u}{u + 1}\, du}{2}$
      
      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: $\frac{u^{3}}{6} - \frac{u^{2}}{4}$
Now substitute $u$ back in:

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

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

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

The answer is:

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

The answer (Indefinite) [src]

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

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

        ___
  2   \/ 3 
- - + -----
  3     2

$$- \frac{2}{3} + \frac{\sqrt{3}}{2}$$

        ___
  2   \/ 3 
- - + -----
  3     2

$$- \frac{2}{3} + \frac{\sqrt{3}}{2}$$

-2/3 + sqrt(3)/2

Numerical answer [src]

0.199358737117772

0.199358737117772

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2