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Solving integrals/ (ln(3x-1))/(3x-1)

Integral of (ln(3x-1))/(3x-1) dx

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The solution

You have entered [src] 
  1                
  /                
 |                 
 |  log(3*x - 1)   
 |  ------------ dx
 |    3*x - 1      
 |                 
/                  
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∫131log⁡(3x−1)3x−1 dx\int\limits_{\frac{1}{3}}^{1} \frac{\log{\left(3 x - 1 \right)}}{3 x - 1}\, dx 
Integral(log(3*x - 1)/(3*x - 1), (x, 1/3, 1))
Detail solution

  1. Let u=3x−1u = 3 x - 1.

    Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

    ∫log⁡(u)3u du\int \frac{\log{\left(u \right)}}{3 u}\, du

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

      ∫log⁡(u)u du=∫log⁡(u)u du3\int \frac{\log{\left(u \right)}}{u}\, du = \frac{\int \frac{\log{\left(u \right)}}{u}\, du}{3}

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=1uu = \frac{1}{u}.

          Then let du=−duu2du = - \frac{du}{u^{2}} and substitute −du- du:

          ∫(−log⁡(1u)u) du\int \left(- \frac{\log{\left(\frac{1}{u} \right)}}{u}\right)\, du

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

            ∫log⁡(1u)u du=−∫log⁡(1u)u du\int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du

            1. Let u=log⁡(1u)u = \log{\left(\frac{1}{u} \right)}.

              Then let du=−duudu = - \frac{du}{u} and substitute −du- du:

              ∫(−u) du\int \left(- u\right)\, du

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

                ∫u du=−∫u du\int u\, du = - \int u\, du

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                  ∫u du=u22\int u\, du = \frac{u^{2}}{2}

                So, the result is: −u22- \frac{u^{2}}{2}

              Now substitute uu back in:

              −log⁡(1u)22- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

            So, the result is: log⁡(1u)22\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

          Now substitute uu back in:

          log⁡(u)22\frac{\log{\left(u \right)}^{2}}{2}

        Method #2

        1. Let u=log⁡(u)u = \log{\left(u \right)}.

          Then let du=duudu = \frac{du}{u} and substitute dudu:

          ∫u du\int u\, du

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

            ∫u du=u22\int u\, du = \frac{u^{2}}{2}

          Now substitute uu back in:

          log⁡(u)22\frac{\log{\left(u \right)}^{2}}{2}

      So, the result is: log⁡(u)26\frac{\log{\left(u \right)}^{2}}{6}

    Now substitute uu back in:

    log⁡(3x−1)26\frac{\log{\left(3 x - 1 \right)}^{2}}{6}

  2. Now simplify:

    log⁡(3x−1)26\frac{\log{\left(3 x - 1 \right)}^{2}}{6}

  3. Add the constant of integration:

    log⁡(3x−1)26+constant\frac{\log{\left(3 x - 1 \right)}^{2}}{6}+ \mathrm{constant}

The answer is:

log⁡(3x−1)26+constant\frac{\log{\left(3 x - 1 \right)}^{2}}{6}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
 |                          2         
 | log(3*x - 1)          log (3*x - 1)
 | ------------ dx = C + -------------
 |   3*x - 1                   6      
 |                                    
/
∫log⁡(3x−1)3x−1 dx=C+log⁡(3x−1)26\int \frac{\log{\left(3 x - 1 \right)}}{3 x - 1}\, dx = C + \frac{\log{\left(3 x - 1 \right)}^{2}}{6}
Simplify
The answer [src] 
-oo
−∞-\infty 
=
= 
-oo
−∞-\infty 
-oo
Numerical answer [src] 
(-283.773849354205 - 1.68049929284057j)
(-283.773849354205 - 1.68049929284057j)

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

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