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Solving integrals/ (3ln^3x)/(x)dx

Integral of (3ln^3x)/(x)dx dx

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The solution

You have entered [src] 
  1             
  /             
 |              
 |       3      
 |  3*log (x)   
 |  --------- dx
 |      x       
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∫013log⁡(x)3x dx\int\limits_{0}^{1} \frac{3 \log{\left(x \right)}^{3}}{x}\, dx 
Integral((3*log(x)^3)/x, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

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

      ∫3u3 du\int 3 u^{3}\, du

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

        ∫u3 du=3∫u3 du\int u^{3}\, du = 3 \int u^{3}\, du

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

          ∫u3 du=u44\int u^{3}\, du = \frac{u^{4}}{4}

        So, the result is: 3u44\frac{3 u^{4}}{4}

      Now substitute uu back in:

      3log⁡(x)44\frac{3 \log{\left(x \right)}^{4}}{4}

    Method #2

    1. Let u=1xu = \frac{1}{x}.

      Then let du=−dxx2du = - \frac{dx}{x^{2}} and substitute −3du- 3 du:

      ∫(−3log⁡(1u)3u) du\int \left(- \frac{3 \log{\left(\frac{1}{u} \right)}^{3}}{u}\right)\, du

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

        ∫log⁡(1u)3u du=−3∫log⁡(1u)3u du\int \frac{\log{\left(\frac{1}{u} \right)}^{3}}{u}\, du = - 3 \int \frac{\log{\left(\frac{1}{u} \right)}^{3}}{u}\, du

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

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

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

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

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

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

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

              ∫u3 du\int u^{3}\, du

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

                ∫u3 du=u44\int u^{3}\, du = \frac{u^{4}}{4}

              Now substitute uu back in:

              log⁡(u)44\frac{\log{\left(u \right)}^{4}}{4}

            So, the result is: −log⁡(u)44- \frac{\log{\left(u \right)}^{4}}{4}

          Now substitute uu back in:

          −log⁡(u)44- \frac{\log{\left(u \right)}^{4}}{4}

        So, the result is: 3log⁡(u)44\frac{3 \log{\left(u \right)}^{4}}{4}

      Now substitute uu back in:

      3log⁡(x)44\frac{3 \log{\left(x \right)}^{4}}{4}

  2. Add the constant of integration:

    3log⁡(x)44+constant\frac{3 \log{\left(x \right)}^{4}}{4}+ \mathrm{constant}

The answer is:

3log⁡(x)44+constant\frac{3 \log{\left(x \right)}^{4}}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                            
 |                             
 |      3                  4   
 | 3*log (x)          3*log (x)
 | --------- dx = C + ---------
 |     x                  4    
 |                             
/
∫3log⁡(x)3x dx=C+3log⁡(x)44\int \frac{3 \log{\left(x \right)}^{3}}{x}\, dx = C + \frac{3 \log{\left(x \right)}^{4}}{4}
Simplify
The answer [src] 
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−∞-\infty 
=
= 
-oo
−∞-\infty 
-oo
Numerical answer [src] 
-2833909.70478493
-2833909.70478493

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

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