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Integral of z
Identical expressions
sqrt(x)*(lnx^ two)*x
square root of (x) multiply by (lnx squared ) multiply by x
square root of (x) multiply by (lnx to the power of two) multiply by x
√(x)*(lnx^2)*x
sqrt(x)*(lnx2)*x
sqrtx*lnx2*x
sqrt(x)*(lnx²)*x
sqrt(x)*(lnx to the power of 2)*x
sqrt(x)(lnx^2)x
sqrt(x)(lnx2)x
sqrtxlnx2x
sqrtxlnx^2x
sqrt(x)*(lnx^2)*xdx

Integral of sqrt(x)(lnx^2)x dx

The solution

You have entered [src]

  1                   
  /                   
 |                    
 |    ___    2        
 |  \/ x *log (x)*x dx
 |                    
/                     
0

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

Integral(sqrt(x)*log(x)^2*x, (x, 0, 1))

Detail solution

Let $u = \log{\left(x \right)}$ .

Then let $du = \frac{dx}{x}$ and substitute $du$ :

$\int u^{2} e^{\frac{5 u}{2}}\, du$
1. 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 $\operatorname{dv}{\left(u \right)} = e^{\frac{5 u}{2}}$ .
  
  Then $\operatorname{du}{\left(u \right)} = 2 u$ .
  
  To find $v{\left(u \right)}$ :
  1. There are multiple ways to do this integral.
    Method #1
    1. Let $u = \frac{5 u}{2}$ .
      
      Then let $du = \frac{5 du}{2}$ and substitute $\frac{2 du}{5}$ :
      
      $\int \frac{4 e^{u}}{25}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 e^{u}}{5}\, du = \frac{2 \int e^{u}\, du}{5}$
      
      The integral of the exponential function is itself.
      
      $\int e^{u}\, du = e^{u}$
      
      So, the result is: $\frac{2 e^{u}}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{2 e^{\frac{5 u}{2}}}{5}$
    Method #2
    1. Let $u = e^{\frac{5 u}{2}}$ .
      
      Then let $du = \frac{5 e^{\frac{5 u}{2}} du}{2}$ and substitute $\frac{2 du}{5}$ :
      
      $\int \frac{4}{25}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2}{5}\, du = \frac{2 \int 1\, du}{5}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int 1\, du = u$
      
      So, the result is: $\frac{2 u}{5}$
      
      Now substitute $u$ back in:
      
      $\frac{2 e^{\frac{5 u}{2}}}{5}$
  Now evaluate the sub-integral.
2. Use integration by parts:
  
  $\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$
  
  Let $u{\left(u \right)} = \frac{4 u}{5}$ and let $\operatorname{dv}{\left(u \right)} = e^{\frac{5 u}{2}}$ .
  
  Then $\operatorname{du}{\left(u \right)} = \frac{4}{5}$ .
  
  To find $v{\left(u \right)}$ :
  1. Let $u = \frac{5 u}{2}$ .
    
    Then let $du = \frac{5 du}{2}$ and substitute $\frac{2 du}{5}$ :
    
    $\int \frac{4 e^{u}}{25}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 e^{u}}{5}\, du = \frac{2 \int e^{u}\, du}{5}$
      1. The integral of the exponential function is itself.
        
        $\int e^{u}\, du = e^{u}$
      So, the result is: $\frac{2 e^{u}}{5}$
    Now substitute $u$ back in:
    
    $\frac{2 e^{\frac{5 u}{2}}}{5}$
  Now evaluate the sub-integral.
3. The integral of a constant times a function is the constant times the integral of the function:
  
  $\int \frac{8 e^{\frac{5 u}{2}}}{25}\, du = \frac{8 \int e^{\frac{5 u}{2}}\, du}{25}$
  1. Let $u = \frac{5 u}{2}$ .
    
    Then let $du = \frac{5 du}{2}$ and substitute $\frac{2 du}{5}$ :
    
    $\int \frac{4 e^{u}}{25}\, du$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{2 e^{u}}{5}\, du = \frac{2 \int e^{u}\, du}{5}$
      1. The integral of the exponential function is itself.
        
        $\int e^{u}\, du = e^{u}$
      So, the result is: $\frac{2 e^{u}}{5}$
    Now substitute $u$ back in:
    
    $\frac{2 e^{\frac{5 u}{2}}}{5}$
  So, the result is: $\frac{16 e^{\frac{5 u}{2}}}{125}$
Now substitute $u$ back in:

$\frac{2 x^{\frac{5}{2}} \log{\left(x \right)}^{2}}{5} - \frac{8 x^{\frac{5}{2}} \log{\left(x \right)}}{25} + \frac{16 x^{\frac{5}{2}}}{125}$
Now simplify:

$\frac{2 x^{\frac{5}{2}} \cdot \left(25 \log{\left(x \right)}^{2} - 20 \log{\left(x \right)} + 8\right)}{125}$
Add the constant of integration:

$\frac{2 x^{\frac{5}{2}} \cdot \left(25 \log{\left(x \right)}^{2} - 20 \log{\left(x \right)} + 8\right)}{125}+ \mathrm{constant}$

The answer is:

$\frac{2 x^{\frac{5}{2}} \cdot \left(25 \log{\left(x \right)}^{2} - 20 \log{\left(x \right)} + 8\right)}{125}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                                                                 
 |                              5/2      5/2             5/2    2   
 |   ___    2               16*x      8*x   *log(x)   2*x   *log (x)
 | \/ x *log (x)*x dx = C + ------- - ------------- + --------------
 |                            125           25              5       
/

$${{8\,x^{{{5}\over{2}}}\,\left({{25\,\left(\log x\right)^2}\over{4}} -5\,\log x+2\right)}\over{125}}$$

Simplify

The answer [src]

 16
---
125

$${{16}\over{125}}$$

 16
---
125

$$\frac{16}{125}$$

Упростить

Numerical answer [src]

0.128

0.128

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

Other calculators

How to use it?

Identical expressions

Integral of sqrt(x)(lnx^2)x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Other calculators

How to use it?

Identical expressions

Integral of sqrt(x)*(lnx^2)*x dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2

Integral of sqrt(x)(lnx^2)x dx