Mister Exam

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Identical expressions
(2x- one)(2x- one)^(one / three)
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(2x-1)(2x-1)^(1/3)dx
Similar expressions
(2x+1)(2x-1)^(1/3)
(2x-1)(2x+1)^(1/3)

Integral of (2x-1)(2x-1)^(1/3) dx

The solution

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  1                         
  /                         
 |                          
 |            3 _________   
 |  (2*x - 1)*\/ 2*x - 1  dx
 |                          
/                           
0

$$\int\limits_{0}^{1} \sqrt[3]{2 x - 1} \left(2 x - 1\right)\, dx$$

Integral((2*x - 1)*(2*x - 1)^(1/3), (x, 0, 1))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = 2 x - 1$ .
  
  Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
  
  $\int \frac{u^{\frac{4}{3}}}{2}\, du$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int u^{\frac{4}{3}}\, du = \frac{\int u^{\frac{4}{3}}\, du}{2}$
    1. The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{\frac{4}{3}}\, du = \frac{3 u^{\frac{7}{3}}}{7}$
    So, the result is: $\frac{3 u^{\frac{7}{3}}}{14}$
  Now substitute $u$ back in:
  
  $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}$
Method #2
1. Rewrite the integrand:
  
  $\sqrt[3]{2 x - 1} \left(2 x - 1\right) = 2 x \sqrt[3]{2 x - 1} - \sqrt[3]{2 x - 1}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x \sqrt[3]{2 x - 1}\, dx = 2 \int x \sqrt[3]{2 x - 1}\, dx$
    1. Let $u = \sqrt[3]{2 x - 1}$ .
      
      Then let $du = \frac{2 dx}{3 \left(2 x - 1\right)^{\frac{2}{3}}}$ and substitute $du$ :
      
      $\int \left(- \frac{3 u^{3}}{4} + 3 \left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2} - \frac{3}{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 u^{3}}{4}\right)\, du = - \frac{3 \int u^{3}\, du}{4}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $- \frac{3 u^{4}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2}\, du = 3 \int \left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2}\, du$
      
      Rewrite the integrand:
      
      $\left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2} = \frac{u^{6}}{4} + \frac{u^{3}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{6}}{4}\, du = \frac{\int u^{6}\, du}{4}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $\frac{u^{7}}{28}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{3}}{2}\, du = \frac{\int u^{3}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $\frac{u^{4}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{u^{7}}{28} + \frac{u^{4}}{8} + \frac{u}{4}$
      
      So, the result is: $\frac{3 u^{7}}{28} + \frac{3 u^{4}}{8} + \frac{3 u}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- \frac{3}{4}\right)\, du = - \frac{3 u}{4}$
      
      The result is: $\frac{3 u^{7}}{28} + \frac{3 u^{4}}{16}$
      
      Now substitute $u$ back in:
      
      $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{28} + \frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{16}$
    So, the result is: $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14} + \frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sqrt[3]{2 x - 1}\right)\, dx = - \int \sqrt[3]{2 x - 1}\, dx$
    1. Let $u = 2 x - 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sqrt[3]{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sqrt[3]{u}\, du = \frac{\int \sqrt[3]{u}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \sqrt[3]{u}\, du = \frac{3 u^{\frac{4}{3}}}{4}$
      
      So, the result is: $\frac{3 u^{\frac{4}{3}}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{8}$
    So, the result is: $- \frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{8}$
  The result is: $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}$
Method #3
1. Rewrite the integrand:
  
  $\sqrt[3]{2 x - 1} \left(2 x - 1\right) = 2 x \sqrt[3]{2 x - 1} - \sqrt[3]{2 x - 1}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int 2 x \sqrt[3]{2 x - 1}\, dx = 2 \int x \sqrt[3]{2 x - 1}\, dx$
    1. Let $u = \sqrt[3]{2 x - 1}$ .
      
      Then let $du = \frac{2 dx}{3 \left(2 x - 1\right)^{\frac{2}{3}}}$ and substitute $du$ :
      
      $\int \left(- \frac{3 u^{3}}{4} + 3 \left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2} - \frac{3}{4}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{3 u^{3}}{4}\right)\, du = - \frac{3 \int u^{3}\, du}{4}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $- \frac{3 u^{4}}{16}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 \left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2}\, du = 3 \int \left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2}\, du$
      
      Rewrite the integrand:
      
      $\left(\frac{u^{3}}{2} + \frac{1}{2}\right)^{2} = \frac{u^{6}}{4} + \frac{u^{3}}{2} + \frac{1}{4}$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{6}}{4}\, du = \frac{\int u^{6}\, du}{4}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      So, the result is: $\frac{u^{7}}{28}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{u^{3}}{2}\, du = \frac{\int u^{3}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{3}\, du = \frac{u^{4}}{4}$
      
      So, the result is: $\frac{u^{4}}{8}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{4}\, du = \frac{u}{4}$
      
      The result is: $\frac{u^{7}}{28} + \frac{u^{4}}{8} + \frac{u}{4}$
      
      So, the result is: $\frac{3 u^{7}}{28} + \frac{3 u^{4}}{8} + \frac{3 u}{4}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- \frac{3}{4}\right)\, du = - \frac{3 u}{4}$
      
      The result is: $\frac{3 u^{7}}{28} + \frac{3 u^{4}}{16}$
      
      Now substitute $u$ back in:
      
      $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{28} + \frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{16}$
    So, the result is: $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14} + \frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{8}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \sqrt[3]{2 x - 1}\right)\, dx = - \int \sqrt[3]{2 x - 1}\, dx$
    1. Let $u = 2 x - 1$ .
      
      Then let $du = 2 dx$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{\sqrt[3]{u}}{2}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \sqrt[3]{u}\, du = \frac{\int \sqrt[3]{u}\, du}{2}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \sqrt[3]{u}\, du = \frac{3 u^{\frac{4}{3}}}{4}$
      
      So, the result is: $\frac{3 u^{\frac{4}{3}}}{8}$
      
      Now substitute $u$ back in:
      
      $\frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{8}$
    So, the result is: $- \frac{3 \left(2 x - 1\right)^{\frac{4}{3}}}{8}$
  The result is: $\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}$
Now simplify:

$\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}$
Add the constant of integration:

$\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}+ \mathrm{constant}$

The answer is:

$\frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                             
 |                                           7/3
 |           3 _________          3*(2*x - 1)   
 | (2*x - 1)*\/ 2*x - 1  dx = C + --------------
 |                                      14      
/

$$\int \sqrt[3]{2 x - 1} \left(2 x - 1\right)\, dx = C + \frac{3 \left(2 x - 1\right)^{\frac{7}{3}}}{14}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

       3 ____
3    3*\/ -1 
-- - --------
14      14

$$\frac{3}{14} - \frac{3 \sqrt[3]{-1}}{14}$$

       3 ____
3    3*\/ -1 
-- - --------
14      14

$$\frac{3}{14} - \frac{3 \sqrt[3]{-1}}{14}$$

3/14 - 3*(-1)^(1/3)/14

Numerical answer [src]

(0.107141103653807 - 0.185573835107398j)

(0.107141103653807 - 0.185573835107398j)

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

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Identical expressions

Similar expressions

Integral of (2x-1)(2x-1)^(1/3) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3