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Integral of (lnxdx)/(x(1+lnx)^1/2) dx

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  1                    
  /                    
 |                     
 |       log(x)        
 |  ---------------- dx
 |      ____________   
 |  x*\/ 1 + log(x)    
 |                     
/                      
0                      
01log(x)xlog(x)+1dx\int\limits_{0}^{1} \frac{\log{\left(x \right)}}{x \sqrt{\log{\left(x \right)} + 1}}\, dx
Integral(log(x)/((x*sqrt(1 + log(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 dudu:

      uu+1du\int \frac{u}{\sqrt{u + 1}}\, du

      1. Let u=1u+1u = \frac{1}{\sqrt{u + 1}}.

        Then let du=du2(u+1)32du = - \frac{du}{2 \left(u + 1\right)^{\frac{3}{2}}} and substitute dudu:

        (2(1+1u2)2+22u2)du\int \left(- 2 \left(-1 + \frac{1}{u^{2}}\right)^{2} + 2 - \frac{2}{u^{2}}\right)\, du

        1. Integrate term-by-term:

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

            (2(1+1u2)2)du=2(1+1u2)2du\int \left(- 2 \left(-1 + \frac{1}{u^{2}}\right)^{2}\right)\, du = - 2 \int \left(-1 + \frac{1}{u^{2}}\right)^{2}\, du

            1. There are multiple ways to do this integral.

              Method #1

              1. Rewrite the integrand:

                (1+1u2)2=12u2+1u4\left(-1 + \frac{1}{u^{2}}\right)^{2} = 1 - \frac{2}{u^{2}} + \frac{1}{u^{4}}

              2. Integrate term-by-term:

                1. The integral of a constant is the constant times the variable of integration:

                  1du=u\int 1\, du = u

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

                  (2u2)du=21u2du\int \left(- \frac{2}{u^{2}}\right)\, du = - 2 \int \frac{1}{u^{2}}\, du

                  1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                    1u2du=1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

                  So, the result is: 2u\frac{2}{u}

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                  1u4du=13u3\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}

                The result is: u+2u13u3u + \frac{2}{u} - \frac{1}{3 u^{3}}

              Method #2

              1. Rewrite the integrand:

                (1+1u2)2=u42u2+1u4\left(-1 + \frac{1}{u^{2}}\right)^{2} = \frac{u^{4} - 2 u^{2} + 1}{u^{4}}

              2. Rewrite the integrand:

                u42u2+1u4=12u2+1u4\frac{u^{4} - 2 u^{2} + 1}{u^{4}} = 1 - \frac{2}{u^{2}} + \frac{1}{u^{4}}

              3. Integrate term-by-term:

                1. The integral of a constant is the constant times the variable of integration:

                  1du=u\int 1\, du = u

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

                  (2u2)du=21u2du\int \left(- \frac{2}{u^{2}}\right)\, du = - 2 \int \frac{1}{u^{2}}\, du

                  1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                    1u2du=1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

                  So, the result is: 2u\frac{2}{u}

                1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                  1u4du=13u3\int \frac{1}{u^{4}}\, du = - \frac{1}{3 u^{3}}

                The result is: u+2u13u3u + \frac{2}{u} - \frac{1}{3 u^{3}}

            So, the result is: 2u4u+23u3- 2 u - \frac{4}{u} + \frac{2}{3 u^{3}}

          1. The integral of a constant is the constant times the variable of integration:

            2du=2u\int 2\, du = 2 u

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

            (2u2)du=21u2du\int \left(- \frac{2}{u^{2}}\right)\, du = - 2 \int \frac{1}{u^{2}}\, du

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              1u2du=1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

            So, the result is: 2u\frac{2}{u}

          The result is: 2u+23u3- \frac{2}{u} + \frac{2}{3 u^{3}}

        Now substitute uu back in:

        2(u+1)3232u+1\frac{2 \left(u + 1\right)^{\frac{3}{2}}}{3} - 2 \sqrt{u + 1}

      Now substitute uu back in:

      2(log(x)+1)3232log(x)+1\frac{2 \left(\log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3} - 2 \sqrt{\log{\left(x \right)} + 1}

    Method #2

    1. Use integration by parts:

      udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=log(x)u{\left(x \right)} = \log{\left(x \right)} and let dv(x)=1xlog(x)+1\operatorname{dv}{\left(x \right)} = \frac{1}{x \sqrt{\log{\left(x \right)} + 1}}.

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

      To find v(x)v{\left(x \right)}:

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

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

        (1ulog(1u)+1)du\int \left(- \frac{1}{u \sqrt{\log{\left(\frac{1}{u} \right)} + 1}}\right)\, du

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

          1ulog(1u)+1du=1ulog(1u)+1du\int \frac{1}{u \sqrt{\log{\left(\frac{1}{u} \right)} + 1}}\, du = - \int \frac{1}{u \sqrt{\log{\left(\frac{1}{u} \right)} + 1}}\, du

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

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

            (1u)du\int \left(- \frac{1}{\sqrt{u}}\right)\, du

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

              1udu=1udu\int \frac{1}{\sqrt{u}}\, du = - \int \frac{1}{\sqrt{u}}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                1udu=2u\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}

              So, the result is: 2u- 2 \sqrt{u}

            Now substitute uu back in:

            2log(1u)+1- 2 \sqrt{\log{\left(\frac{1}{u} \right)} + 1}

          So, the result is: 2log(1u)+12 \sqrt{\log{\left(\frac{1}{u} \right)} + 1}

        Now substitute uu back in:

        2log(x)+12 \sqrt{\log{\left(x \right)} + 1}

      Now evaluate the sub-integral.

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

      2log(x)+1xdx=2log(x)+1xdx\int \frac{2 \sqrt{\log{\left(x \right)} + 1}}{x}\, dx = 2 \int \frac{\sqrt{\log{\left(x \right)} + 1}}{x}\, dx

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

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

        (log(1u)+1u)du\int \left(- \frac{\sqrt{\log{\left(\frac{1}{u} \right)} + 1}}{u}\right)\, du

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

          log(1u)+1udu=log(1u)+1udu\int \frac{\sqrt{\log{\left(\frac{1}{u} \right)} + 1}}{u}\, du = - \int \frac{\sqrt{\log{\left(\frac{1}{u} \right)} + 1}}{u}\, du

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

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

            (u)du\int \left(- \sqrt{u}\right)\, du

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

              udu=udu\int \sqrt{u}\, du = - \int \sqrt{u}\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

                udu=2u323\int \sqrt{u}\, du = \frac{2 u^{\frac{3}{2}}}{3}

              So, the result is: 2u323- \frac{2 u^{\frac{3}{2}}}{3}

            Now substitute uu back in:

            2(log(1u)+1)323- \frac{2 \left(\log{\left(\frac{1}{u} \right)} + 1\right)^{\frac{3}{2}}}{3}

          So, the result is: 2(log(1u)+1)323\frac{2 \left(\log{\left(\frac{1}{u} \right)} + 1\right)^{\frac{3}{2}}}{3}

        Now substitute uu back in:

        2(log(x)+1)323\frac{2 \left(\log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3}

      So, the result is: 4(log(x)+1)323\frac{4 \left(\log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3}

  2. Now simplify:

    2(log(x)2)log(x)+13\frac{2 \left(\log{\left(x \right)} - 2\right) \sqrt{\log{\left(x \right)} + 1}}{3}

  3. Add the constant of integration:

    2(log(x)2)log(x)+13+constant\frac{2 \left(\log{\left(x \right)} - 2\right) \sqrt{\log{\left(x \right)} + 1}}{3}+ \mathrm{constant}


The answer is:

2(log(x)2)log(x)+13+constant\frac{2 \left(\log{\left(x \right)} - 2\right) \sqrt{\log{\left(x \right)} + 1}}{3}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                              
 |                                                            3/2
 |      log(x)                   ____________   2*(1 + log(x))   
 | ---------------- dx = C - 2*\/ 1 + log(x)  + -----------------
 |     ____________                                     3        
 | x*\/ 1 + log(x)                                               
 |                                                               
/                                                                
log(x)xlog(x)+1dx=C+2(log(x)+1)3232log(x)+1\int \frac{\log{\left(x \right)}}{x \sqrt{\log{\left(x \right)} + 1}}\, dx = C + \frac{2 \left(\log{\left(x \right)} + 1\right)^{\frac{3}{2}}}{3} - 2 \sqrt{\log{\left(x \right)} + 1}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-200200
The answer [src]
-4/3 + oo*I
43+i- \frac{4}{3} + \infty i
=
=
-4/3 + oo*I
43+i- \frac{4}{3} + \infty i
-4/3 + oo*i
Numerical answer [src]
(-1.0818887836703 + 202.533724926781j)
(-1.0818887836703 + 202.533724926781j)

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