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Solving integrals/ dx/xln4x

Integral of dx/xln4x dx

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  1            
  /            
 |             
 |  log(4*x)   
 |  -------- dx
 |     x       
 |             
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0
∫01log⁡(4x)x dx\int\limits_{0}^{1} \frac{\log{\left(4 x \right)}}{x}\, dx 
Integral(log(4*x)/x, (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

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

      ∫(−log⁡(1u)+2log⁡(2)u) du\int \left(- \frac{\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}}{u}\right)\, du

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

        ∫log⁡(1u)+2log⁡(2)u du=−∫log⁡(1u)+2log⁡(2)u du\int \frac{\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}}{u}\, du

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

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

          ∫(−log⁡(u)+2log⁡(2)u) du\int \left(- \frac{\log{\left(u \right)} + 2 \log{\left(2 \right)}}{u}\right)\, du

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

            ∫log⁡(u)+2log⁡(2)u du=−∫log⁡(u)+2log⁡(2)u du\int \frac{\log{\left(u \right)} + 2 \log{\left(2 \right)}}{u}\, du = - \int \frac{\log{\left(u \right)} + 2 \log{\left(2 \right)}}{u}\, du

            1. Let u=log⁡(u)+2log⁡(2)u = \log{\left(u \right)} + 2 \log{\left(2 \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)+2log⁡(2))22\frac{\left(\log{\left(u \right)} + 2 \log{\left(2 \right)}\right)^{2}}{2}

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

          Now substitute uu back in:

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

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

      Now substitute uu back in:

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

    Method #2

    1. Rewrite the integrand:

      log⁡(4x)x=log⁡(x)+2log⁡(2)x\frac{\log{\left(4 x \right)}}{x} = \frac{\log{\left(x \right)} + 2 \log{\left(2 \right)}}{x}

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

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

      ∫(−log⁡(1u)+2log⁡(2)u) du\int \left(- \frac{\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}}{u}\right)\, du

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

        ∫log⁡(1u)+2log⁡(2)u du=−∫log⁡(1u)+2log⁡(2)u du\int \frac{\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}}{u}\, du = - \int \frac{\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}}{u}\, du

        1. Let u=log⁡(1u)+2log⁡(2)u = \log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \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)+2log⁡(2))22- \frac{\left(\log{\left(\frac{1}{u} \right)} + 2 \log{\left(2 \right)}\right)^{2}}{2}

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

      Now substitute uu back in:

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

    Method #3

    1. Rewrite the integrand:

      log⁡(4x)x=log⁡(x)x+2log⁡(2)x\frac{\log{\left(4 x \right)}}{x} = \frac{\log{\left(x \right)}}{x} + \frac{2 \log{\left(2 \right)}}{x}

    2. Integrate term-by-term:

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

        Then let du=−dxx2du = - \frac{dx}{x^{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⁡(x)22\frac{\log{\left(x \right)}^{2}}{2}

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

        ∫2log⁡(2)x dx=2log⁡(2)∫1x dx\int \frac{2 \log{\left(2 \right)}}{x}\, dx = 2 \log{\left(2 \right)} \int \frac{1}{x}\, dx

        1. The integral of 1x\frac{1}{x} is log⁡(x)\log{\left(x \right)}.

        So, the result is: 2log⁡(2)log⁡(x)2 \log{\left(2 \right)} \log{\left(x \right)}

      The result is: log⁡(x)22+2log⁡(2)log⁡(x)\frac{\log{\left(x \right)}^{2}}{2} + 2 \log{\left(2 \right)} \log{\left(x \right)}

  2. Now simplify:

    log⁡(4x)22\frac{\log{\left(4 x \right)}^{2}}{2}

  3. Add the constant of integration:

    log⁡(4x)22+constant\frac{\log{\left(4 x \right)}^{2}}{2}+ \mathrm{constant}

The answer is:

log⁡(4x)22+constant\frac{\log{\left(4 x \right)}^{2}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                      
 |                                      2
 | log(4*x)          (2*log(2) + log(x)) 
 | -------- dx = C + --------------------
 |    x                       2          
 |                                       
/
∫log⁡(4x)x dx=C+(log⁡(x)+2log⁡(2))22\int \frac{\log{\left(4 x \right)}}{x}\, dx = C + \frac{\left(\log{\left(x \right)} + 2 \log{\left(2 \right)}\right)^{2}}{2}
Simplify
The answer [src] 
-oo
−∞-\infty 
=
= 
-oo
−∞-\infty 
-oo
Numerical answer [src] 
-910.841526560512
-910.841526560512

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

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