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Integral of d{x}:
Integral of cotx
Integral of xlnxdx
Integral of tan⁵x
Integral of ln2x
Identical expressions
(3x+ two)*ln(x)^ two
(3x plus 2) multiply by ln(x) squared
(3x plus two) multiply by ln(x) to the power of two
(3x+2)*ln(x)2
3x+2*lnx2
(3x+2)*ln(x)²
(3x+2)*ln(x) to the power of 2
(3x+2)ln(x)^2
(3x+2)ln(x)2
3x+2lnx2
3x+2lnx^2
(3x+2)*ln(x)^2dx
Similar expressions
(3x-2)*ln(x)^2

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

The solution

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 |  (3*x + 2)*log (x) dx
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0

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

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

Detail solution

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

$\frac{x \left(6 x \log{\left(x \right)}^{2} - 6 x \log{\left(x \right)} + 3 x + 8 \log{\left(x \right)}^{2} - 16 \log{\left(x \right)} + 16\right)}{4}$
Add the constant of integration:

$\frac{x \left(6 x \log{\left(x \right)}^{2} - 6 x \log{\left(x \right)} + 3 x + 8 \log{\left(x \right)}^{2} - 16 \log{\left(x \right)} + 16\right)}{4}+ \mathrm{constant}$

The answer is:

$\frac{x \left(6 x \log{\left(x \right)}^{2} - 6 x \log{\left(x \right)} + 3 x + 8 \log{\left(x \right)}^{2} - 16 \log{\left(x \right)} + 16\right)}{4}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                                             
 |                                     2                                 2             2    2   
 |              2                   3*x                        2      3*x *log(x)   3*x *log (x)
 | (3*x + 2)*log (x) dx = C + 4*x + ---- - 4*x*log(x) + 2*x*log (x) - ----------- + ------------
 |                                   4                                     2             2      
/

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

Simplify

The graph

Plot the graph f(x)

The answer [src]

19/4

$$\frac{19}{4}$$

19/4

$$\frac{19}{4}$$

19/4

Numerical answer [src]

4.75

4.75

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

Other calculators

How to use it?

Identical expressions

Similar expressions

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

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Method #3