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(3x+2)/(x+1)^(1/4)dx
Similar expressions
(3x-2)/(x+1)^(1/4)
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Integral of (3x+2)/(x+1)^(1/4) dx

The solution

You have entered [src]

 15             
  /             
 |              
 |   3*x + 2    
 |  --------- dx
 |  4 _______   
 |  \/ x + 1    
 |              
/               
0

$$\int\limits_{0}^{15} \frac{3 x + 2}{\sqrt[4]{x + 1}}\, dx$$

Integral((3*x + 2)/(x + 1)^(1/4), (x, 0, 15))

Detail solution

There are multiple ways to do this integral.
Method #1
1. Let $u = \sqrt[4]{x + 1}$ .
  
  Then let $du = \frac{dx}{4 \left(x + 1\right)^{\frac{3}{4}}}$ and substitute $du$ :
  
  $\int \left(12 u^{2} \left(u^{4} - 1\right) + 8 u^{2}\right)\, du$
  1. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 12 u^{2} \left(u^{4} - 1\right)\, du = 12 \int u^{2} \left(u^{4} - 1\right)\, du$
      
      Rewrite the integrand:
      
      $u^{2} \left(u^{4} - 1\right) = u^{6} - u^{2}$
      
      Integrate term-by-term:
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int u^{6}\, du = \frac{u^{7}}{7}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- u^{2}\right)\, du = - \int u^{2}\, du$
      
      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: $- \frac{u^{3}}{3}$
      
      The result is: $\frac{u^{7}}{7} - \frac{u^{3}}{3}$
      
      So, the result is: $\frac{12 u^{7}}{7} - 4 u^{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 8 u^{2}\, du = 8 \int u^{2}\, du$
      
      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: $\frac{8 u^{3}}{3}$
    The result is: $\frac{12 u^{7}}{7} - \frac{4 u^{3}}{3}$
  Now substitute $u$ back in:
  
  $\frac{12 \left(x + 1\right)^{\frac{7}{4}}}{7} - \frac{4 \left(x + 1\right)^{\frac{3}{4}}}{3}$
Method #2
1. Rewrite the integrand:
  
  $\frac{3 x + 2}{\sqrt[4]{x + 1}} = \frac{3 x}{\sqrt[4]{x + 1}} + \frac{2}{\sqrt[4]{x + 1}}$
2. Integrate term-by-term:
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{3 x}{\sqrt[4]{x + 1}}\, dx = 3 \int \frac{x}{\sqrt[4]{x + 1}}\, dx$
    1. Let $u = \frac{1}{\sqrt[4]{x + 1}}$ .
      
      Then let $du = - \frac{dx}{4 \left(x + 1\right)^{\frac{5}{4}}}$ and substitute $du$ :
      
      $\int \left(- 4 \left(-1 + \frac{1}{u^{4}}\right)^{2} + 4 - \frac{4}{u^{4}}\right)\, du$
      
      Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- 4 \left(-1 + \frac{1}{u^{4}}\right)^{2}\right)\, du = - 4 \int \left(-1 + \frac{1}{u^{4}}\right)^{2}\, du$
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Rewrite the integrand:
      
      $\left(-1 + \frac{1}{u^{4}}\right)^{2} = 1 - \frac{2}{u^{4}} + \frac{1}{u^{8}}$
      
      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:
      
      $\int \left(- \frac{2}{u^{4}}\right)\, du = - 2 \int \frac{1}{u^{4}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$
      
      So, the result is: $\frac{2}{3 u^{3}}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{8}}\, du = - \frac{1}{7 u^{7}}$
      
      The result is: $u + \frac{2}{3 u^{3}} - \frac{1}{7 u^{7}}$
      
      Method #2
      
      Rewrite the integrand:
      
      $\left(-1 + \frac{1}{u^{4}}\right)^{2} = \frac{u^{8} - 2 u^{4} + 1}{u^{8}}$
      
      Rewrite the integrand:
      
      $\frac{u^{8} - 2 u^{4} + 1}{u^{8}} = 1 - \frac{2}{u^{4}} + \frac{1}{u^{8}}$
      
      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:
      
      $\int \left(- \frac{2}{u^{4}}\right)\, du = - 2 \int \frac{1}{u^{4}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$
      
      So, the result is: $\frac{2}{3 u^{3}}$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{8}}\, du = - \frac{1}{7 u^{7}}$
      
      The result is: $u + \frac{2}{3 u^{3}} - \frac{1}{7 u^{7}}$
      
      So, the result is: $- 4 u - \frac{8}{3 u^{3}} + \frac{4}{7 u^{7}}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 4\, du = 4 u$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{4}{u^{4}}\right)\, du = - 4 \int \frac{1}{u^{4}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}$
      
      So, the result is: $\frac{4}{3 u^{3}}$
      
      The result is: $- \frac{4}{3 u^{3}} + \frac{4}{7 u^{7}}$
      
      Now substitute $u$ back in:
      
      $\frac{4 \left(x + 1\right)^{\frac{7}{4}}}{7} - \frac{4 \left(x + 1\right)^{\frac{3}{4}}}{3}$
    So, the result is: $\frac{12 \left(x + 1\right)^{\frac{7}{4}}}{7} - 4 \left(x + 1\right)^{\frac{3}{4}}$
  1. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \frac{2}{\sqrt[4]{x + 1}}\, dx = 2 \int \frac{1}{\sqrt[4]{x + 1}}\, dx$
    1. Let $u = x + 1$ .
      
      Then let $du = dx$ and substitute $du$ :
      
      $\int \frac{1}{\sqrt[4]{u}}\, du$
      
      The integral of $u^{n}$ is $\frac{u^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int \frac{1}{\sqrt[4]{u}}\, du = \frac{4 u^{\frac{3}{4}}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{4 \left(x + 1\right)^{\frac{3}{4}}}{3}$
    So, the result is: $\frac{8 \left(x + 1\right)^{\frac{3}{4}}}{3}$
  The result is: $\frac{12 \left(x + 1\right)^{\frac{7}{4}}}{7} - \frac{4 \left(x + 1\right)^{\frac{3}{4}}}{3}$
Now simplify:

$\frac{4 \left(x + 1\right)^{\frac{3}{4}} \left(9 x + 2\right)}{21}$
Add the constant of integration:

$\frac{4 \left(x + 1\right)^{\frac{3}{4}} \left(9 x + 2\right)}{21}+ \mathrm{constant}$

The answer is:

$\frac{4 \left(x + 1\right)^{\frac{3}{4}} \left(9 x + 2\right)}{21}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                               
 |                             3/4             7/4
 |  3*x + 2           4*(x + 1)      12*(x + 1)   
 | --------- dx = C - ------------ + -------------
 | 4 _______               3               7      
 | \/ x + 1                                       
 |                                                
/

$$\int \frac{3 x + 2}{\sqrt[4]{x + 1}}\, dx = C + \frac{12 \left(x + 1\right)^{\frac{7}{4}}}{7} - \frac{4 \left(x + 1\right)^{\frac{3}{4}}}{3}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

4376
----
 21

$$\frac{4376}{21}$$

4376
----
 21

$$\frac{4376}{21}$$

4376/21

Numerical answer [src]

208.380952380952

208.380952380952

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

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Identical expressions

Similar expressions

Integral of (3x+2)/(x+1)^(1/4) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #1

Method #2