Integral of xln(1-3x) dx

The solution

You have entered [src]

  0                  
  /                  
 |                   
 |  x*log(1 - 3*x) dx
 |                   
/                    
0

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

Integral(x*log(1 - 3*x), (x, 0, 0))

Detail solution

Use integration by parts:

$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$

Let $u{\left(x \right)} = \log{\left(1 - 3 x \right)}$ and let $\operatorname{dv}{\left(x \right)} = x$ .

Then $\operatorname{du}{\left(x \right)} = - \frac{3}{1 - 3 x}$ .

To find $v{\left(x \right)}$ :
1. The integral of $x^{n}$ is $\frac{x^{n + 1}}{n + 1}$ when $n \neq -1$ :
  
  $\int x\, dx = \frac{x^{2}}{2}$
Now evaluate the sub-integral.
The integral of a constant times a function is the constant times the integral of the function:

$\int \left(- \frac{3 x^{2}}{2 \cdot \left(1 - 3 x\right)}\right)\, dx = - \frac{3 \int \frac{x^{2}}{1 - 3 x}\, dx}{2}$
1. There are multiple ways to do this integral.
  Method #1
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{1 - 3 x} = - \frac{x}{3} - \frac{1}{9} - \frac{1}{9 \cdot \left(3 x - 1\right)}$
  2. Integrate term-by-term:
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{x}{3}\right)\, dx = - \frac{\int x\, dx}{3}$
      
      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: $- \frac{x^{2}}{6}$
    1. The integral of a constant is the constant times the variable of integration:
      
      $\int \left(- \frac{1}{9}\right)\, dx = - \frac{x}{9}$
    1. The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \left(- \frac{1}{9 \cdot \left(3 x - 1\right)}\right)\, dx = - \frac{\int \frac{1}{3 x - 1}\, dx}{9}$
      
      Let $u = 3 x - 1$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{1}{9 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{3 u}\, du = \frac{\int \frac{1}{u}\, du}{3}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(3 x - 1 \right)}}{3}$
      
      So, the result is: $- \frac{\log{\left(3 x - 1 \right)}}{27}$
    The result is: $- \frac{x^{2}}{6} - \frac{x}{9} - \frac{\log{\left(3 x - 1 \right)}}{27}$
  Method #2
  1. Rewrite the integrand:
    
    $\frac{x^{2}}{1 - 3 x} = - \frac{x^{2}}{3 x - 1}$
  2. The integral of a constant times a function is the constant times the integral of the function:
    
    $\int \left(- \frac{x^{2}}{3 x - 1}\right)\, dx = - \int \frac{x^{2}}{3 x - 1}\, dx$
    1. Rewrite the integrand:
      
      $\frac{x^{2}}{3 x - 1} = \frac{x}{3} + \frac{1}{9} + \frac{1}{9 \cdot \left(3 x - 1\right)}$
    2. Integrate term-by-term:
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{x}{3}\, dx = \frac{\int x\, dx}{3}$
      
      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: $\frac{x^{2}}{6}$
      
      The integral of a constant is the constant times the variable of integration:
      
      $\int \frac{1}{9}\, dx = \frac{x}{9}$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{9 \cdot \left(3 x - 1\right)}\, dx = \frac{\int \frac{1}{3 x - 1}\, dx}{9}$
      
      Let $u = 3 x - 1$ .
      
      Then let $du = 3 dx$ and substitute $\frac{du}{3}$ :
      
      $\int \frac{1}{9 u}\, du$
      
      The integral of a constant times a function is the constant times the integral of the function:
      
      $\int \frac{1}{3 u}\, du = \frac{\int \frac{1}{u}\, du}{3}$
      
      The integral of $\frac{1}{u}$ is $\log{\left(u \right)}$ .
      
      So, the result is: $\frac{\log{\left(u \right)}}{3}$
      
      Now substitute $u$ back in:
      
      $\frac{\log{\left(3 x - 1 \right)}}{3}$
      
      So, the result is: $\frac{\log{\left(3 x - 1 \right)}}{27}$
      
      The result is: $\frac{x^{2}}{6} + \frac{x}{9} + \frac{\log{\left(3 x - 1 \right)}}{27}$
    So, the result is: $- \frac{x^{2}}{6} - \frac{x}{9} - \frac{\log{\left(3 x - 1 \right)}}{27}$
So, the result is: $\frac{x^{2}}{4} + \frac{x}{6} + \frac{\log{\left(3 x - 1 \right)}}{18}$
Now simplify:

$\frac{x^{2} \log{\left(1 - 3 x \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{6} - \frac{\log{\left(3 x - 1 \right)}}{18}$
Add the constant of integration:

$\frac{x^{2} \log{\left(1 - 3 x \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{6} - \frac{\log{\left(3 x - 1 \right)}}{18}+ \mathrm{constant}$

The answer is:

$\frac{x^{2} \log{\left(1 - 3 x \right)}}{2} - \frac{x^{2}}{4} - \frac{x}{6} - \frac{\log{\left(3 x - 1 \right)}}{18}+ \mathrm{constant}$

The answer (Indefinite) [src]

  /                         2                        2             
 |                         x    x   log(-1 + 3*x)   x *log(1 - 3*x)
 | x*log(1 - 3*x) dx = C - -- - - - ------------- + ---------------
 |                         4    6         18               2       
/

$${{3\,\left(-{{\log \left(3\,x-1\right)}\over{27}}-{{3\,x^2+2\,x }\over{18}}\right)}\over{2}}+{{\log \left(1-3\,x\right)\,x^2}\over{2 }}$$

Simplify

The graph

Plot the graph f(x)

The answer [src]

$$0$$

Упростить

Numerical answer [src]

0.0

0.0

The graph

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

Other calculators

How to use it?

Identical expressions

Similar expressions

Integral of xln(1-3x) dx

Limits of integration:

The graph:

Piecewise:

The solution

Method #1

Method #2