Integral of (3x+2)ln*x*bx dx

You have entered [src]

  1                        
  /                        
 |                         
 |  (3*x + 2)*log(x)*b*x dx
 |                         
/                          
0

$$\int\limits_{0}^{1} b x \left(3 x + 2\right) \log{\left(x \right)}\, dx$$

Integral((3*x + 2)*log(x)*b*x, (x, 0, 1))

Detail solution

The integral of a constant times a function is the constant times the integral of the function:

$\int b x \left(3 x + 2\right) \log{\left(x \right)}\, dx = b \int x \left(3 x + 2\right) \log{\left(x \right)}\, dx$
1. There are multiple ways to do this integral.
  Method #1
  1. Let $u = \log{\left(x \right)}$ .
    
    Then let $du = \frac{dx}{x}$ and substitute $du$ :
    
    $\int \left(3 u e^{3 u} + 2 u e^{2 u}\right)\, du$
    1. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 u e^{3 u}\, du = 3 \int u e^{3 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      There are multiple ways to do this integral.
      
      Method #1
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Method #2
      
      Let $u = e^{3 u}$ .
      
      Then let $du = 3 e^{3 u} du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{1}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{3}\, du = \frac{\int 1\, du}{3}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{u}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
      
      So, the result is: $u e^{3 u} - \frac{e^{3 u}}{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 u e^{2 u}\, du = 2 \int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      So, the result is: $u e^{2 u} - \frac{e^{2 u}}{2}$
      
      The result is: $u e^{3 u} + u e^{2 u} - \frac{e^{3 u}}{3} - \frac{e^{2 u}}{2}$
    Now substitute $u$ back in:
    
    $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
  Method #2
  1. Rewrite the integrand:
    
    $x \left(3 x + 2\right) \log{\left(x \right)} = 3 x^{2} \log{\left(x \right)} + 2 x \log{\left(x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 x^{2} \log{\left(x \right)}\, dx = 3 \int x^{2} \log{\left(x \right)}\, dx$
      
      Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{3 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$
      
      So, the result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 x \log{\left(x \right)}\, dx = 2 \int x \log{\left(x \right)}\, dx$
      
      Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
      
      So, the result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
    The result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
  Method #3
  1. Use integration by parts:
    
    $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
    
    Let $u{\left(x \right)} = \log{\left(x \right)}$ and let $\operatorname{dv}{\left(x \right)} = x \left(3 x + 2\right)$ .
    
    Then $\operatorname{du}{\left(x \right)} = \frac{1}{x}$ .
    
    To find $v{\left(x \right)}$ :
    1. Rewrite the integrand:
      
      $x \left(3 x + 2\right) = 3 x^{2} + 2 x$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 x^{2}\, dx = 3 \int x^{2}\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
      
      So, the result is: $x^{3}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 x\, dx = 2 \int x\, dx$
      
      The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
      
      So, the result is: $x^{2}$
      
      The result is: $x^{3} + x^{2}$
    Now evaluate the sub-integral.
  2. Rewrite the integrand:
    
    $\frac{x^{3} + x^{2}}{x} = x^{2} + x$
  3. Integrate term-by-term:
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x^{2}\, dx = \frac{x^{3}}{3}$
    1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
      
      $\int x\, dx = \frac{x^{2}}{2}$
    The result is: $\frac{x^{3}}{3} + \frac{x^{2}}{2}$
  Method #4
  1. Rewrite the integrand:
    
    $x \left(3 x + 2\right) \log{\left(x \right)} = 3 x^{2} \log{\left(x \right)} + 2 x \log{\left(x \right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 3 x^{2} \log{\left(x \right)}\, dx = 3 \int x^{2} \log{\left(x \right)}\, dx$
      
      Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{3 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{3 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}$
      
      Let $u = 3 u$ .
      
      Then let $du = 3 du$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{e^{u}}{9}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{3}\, du = \frac{\int e^{u}\, du}{3}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{3 u}}{3}$
      
      So, the result is: $\frac{e^{3 u}}{9}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}$
      
      So, the result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int 2 x \log{\left(x \right)}\, dx = 2 \int x \log{\left(x \right)}\, dx$
      
      Let $u = \log{\left(x \right)}$ .
      
      Then let $du = \frac{dx}{x}$ and substitute $du$ :
      
      $\int u e^{2 u}\, du$
      
      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 $\operatorname{dv}{\left(u \right)} = e^{2 u}$ .
      
      Then $\operatorname{du}{\left(u \right)} = 1$ .
      
      To find $v{\left(u \right)}$ :
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      Now evaluate the sub-integral.
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{2 u}}{2}\, du = \frac{\int e^{2 u}\, du}{2}$
      
      Let $u = 2 u$ .
      
      Then let $du = 2 du$ and substitute $\frac{du}{2}$ :
      
      $\int \frac{e^{u}}{4}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{e^{u}}{2}\, du = \frac{\int e^{u}\, du}{2}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{e^{u}}{2}$
      
      Now substitute $u$ back in:
      
      $\frac{e^{2 u}}{2}$
      
      So, the result is: $\frac{e^{2 u}}{4}$
      
      Now substitute $u$ back in:
      
      $\frac{x^{2} \log{\left(x \right)}}{2} - \frac{x^{2}}{4}$
      
      So, the result is: $x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
    The result is: $x^{3} \log{\left(x \right)} - \frac{x^{3}}{3} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}$
So, the result is: $b \left(x^{3} \log{\left(x \right)} - \frac{x^{3}}{3} + x^{2} \log{\left(x \right)} - \frac{x^{2}}{2}\right)$
Now simplify:

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

$\frac{b x^{2} \cdot \left(6 x \log{\left(x \right)} - 2 x + 6 \log{\left(x \right)} - 3\right)}{6}+ \mathrm{constant}$

The answer is:

$\frac{b x^{2} \cdot \left(6 x \log{\left(x \right)} - 2 x + 6 \log{\left(x \right)} - 3\right)}{6}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                /   2    3                        \
 |                                 |  x    x     2           3       |
 | (3*x + 2)*log(x)*b*x dx = C + b*|- -- - -- + x *log(x) + x *log(x)|
 |                                 \  2    3                         /
/

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

Simplify

The answer [src]

-5*b
----
 6

$$- \frac{5 b}{6}$$

=

-5*b
----
 6

$$- \frac{5 b}{6}$$

Упростить

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of (3x+2)lnxbx dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #1

Method #2

Method #2

Method #3

Method #4