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Integral of √x*ln^2(x) dx

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01xlog(x)2dx\int\limits_{0}^{1} \sqrt{x} \log{\left(x \right)}^{2}\, dx
Integral(sqrt(x)*log(x)^2, (x, 0, 1))
Detail solution
  1. Let u=log(x)u = \log{\left(x \right)}.

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

    u2e3u2du\int u^{2} e^{\frac{3 u}{2}}\, du

    1. Use integration by parts:

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

      Let u(u)=u2u{\left(u \right)} = u^{2} and let dv(u)=e3u2\operatorname{dv}{\left(u \right)} = e^{\frac{3 u}{2}}.

      Then du(u)=2u\operatorname{du}{\left(u \right)} = 2 u.

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

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=3u2u = \frac{3 u}{2}.

          Then let du=3du2du = \frac{3 du}{2} and substitute 2du3\frac{2 du}{3}:

          4eu9du\int \frac{4 e^{u}}{9}\, du

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

            2eu3du=2eudu3\int \frac{2 e^{u}}{3}\, du = \frac{2 \int e^{u}\, du}{3}

            1. The integral of the exponential function is itself.

              eudu=eu\int e^{u}\, du = e^{u}

            So, the result is: 2eu3\frac{2 e^{u}}{3}

          Now substitute uu back in:

          2e3u23\frac{2 e^{\frac{3 u}{2}}}{3}

        Method #2

        1. Let u=e3u2u = e^{\frac{3 u}{2}}.

          Then let du=3e3u2du2du = \frac{3 e^{\frac{3 u}{2}} du}{2} and substitute 2du3\frac{2 du}{3}:

          49du\int \frac{4}{9}\, du

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

            23du=21du3\int \frac{2}{3}\, du = \frac{2 \int 1\, du}{3}

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

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

            So, the result is: 2u3\frac{2 u}{3}

          Now substitute uu back in:

          2e3u23\frac{2 e^{\frac{3 u}{2}}}{3}

      Now evaluate the sub-integral.

    2. Use integration by parts:

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

      Let u(u)=4u3u{\left(u \right)} = \frac{4 u}{3} and let dv(u)=e3u2\operatorname{dv}{\left(u \right)} = e^{\frac{3 u}{2}}.

      Then du(u)=43\operatorname{du}{\left(u \right)} = \frac{4}{3}.

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

      1. Let u=3u2u = \frac{3 u}{2}.

        Then let du=3du2du = \frac{3 du}{2} and substitute 2du3\frac{2 du}{3}:

        4eu9du\int \frac{4 e^{u}}{9}\, du

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

          2eu3du=2eudu3\int \frac{2 e^{u}}{3}\, du = \frac{2 \int e^{u}\, du}{3}

          1. The integral of the exponential function is itself.

            eudu=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu3\frac{2 e^{u}}{3}

        Now substitute uu back in:

        2e3u23\frac{2 e^{\frac{3 u}{2}}}{3}

      Now evaluate the sub-integral.

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

      8e3u29du=8e3u2du9\int \frac{8 e^{\frac{3 u}{2}}}{9}\, du = \frac{8 \int e^{\frac{3 u}{2}}\, du}{9}

      1. Let u=3u2u = \frac{3 u}{2}.

        Then let du=3du2du = \frac{3 du}{2} and substitute 2du3\frac{2 du}{3}:

        4eu9du\int \frac{4 e^{u}}{9}\, du

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

          2eu3du=2eudu3\int \frac{2 e^{u}}{3}\, du = \frac{2 \int e^{u}\, du}{3}

          1. The integral of the exponential function is itself.

            eudu=eu\int e^{u}\, du = e^{u}

          So, the result is: 2eu3\frac{2 e^{u}}{3}

        Now substitute uu back in:

        2e3u23\frac{2 e^{\frac{3 u}{2}}}{3}

      So, the result is: 16e3u227\frac{16 e^{\frac{3 u}{2}}}{27}

    Now substitute uu back in:

    2x32log(x)238x32log(x)9+16x3227\frac{2 x^{\frac{3}{2}} \log{\left(x \right)}^{2}}{3} - \frac{8 x^{\frac{3}{2}} \log{\left(x \right)}}{9} + \frac{16 x^{\frac{3}{2}}}{27}

  2. Now simplify:

    2x32(9log(x)212log(x)+8)27\frac{2 x^{\frac{3}{2}} \cdot \left(9 \log{\left(x \right)}^{2} - 12 \log{\left(x \right)} + 8\right)}{27}

  3. Add the constant of integration:

    2x32(9log(x)212log(x)+8)27+constant\frac{2 x^{\frac{3}{2}} \cdot \left(9 \log{\left(x \right)}^{2} - 12 \log{\left(x \right)} + 8\right)}{27}+ \mathrm{constant}


The answer is:

2x32(9log(x)212log(x)+8)27+constant\frac{2 x^{\frac{3}{2}} \cdot \left(9 \log{\left(x \right)}^{2} - 12 \log{\left(x \right)} + 8\right)}{27}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                               
 |                            3/2      3/2             3/2    2   
 |   ___    2             16*x      8*x   *log(x)   2*x   *log (x)
 | \/ x *log (x) dx = C + ------- - ------------- + --------------
 |                           27           9               3       
/                                                                 
8x32(9(logx)243logx+2)27{{8\,x^{{{3}\over{2}}}\,\left({{9\,\left(\log x\right)^2}\over{4}}- 3\,\log x+2\right)}\over{27}}
The answer [src]
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1627{{16}\over{27}}
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01{0for1x<1x<1xlog(x)2for1x<1x<12(G4,40,4(32,52,52,132,32,32,0|x)+3G4,40,4(52,52,52,132,32,32,0|x)2)x+2G4,43,1(052,52,5232,32,320|x)xotherwisedx\int\limits_{0}^{1} \begin{cases} 0 & \text{for}\: \frac{1}{x} < 1 \wedge x < 1 \\\sqrt{x} \log{\left(x \right)}^{2} & \text{for}\: \frac{1}{x} < 1 \vee x < 1 \\\frac{2 \left({G_{4, 4}^{0, 4}\left(\begin{matrix} \frac{3}{2}, \frac{5}{2}, \frac{5}{2}, 1 & \\ & \frac{3}{2}, \frac{3}{2}, \frac{3}{2}, 0 \end{matrix} \middle| {x} \right)} + \frac{3 {G_{4, 4}^{0, 4}\left(\begin{matrix} \frac{5}{2}, \frac{5}{2}, \frac{5}{2}, 1 & \\ & \frac{3}{2}, \frac{3}{2}, \frac{3}{2}, 0 \end{matrix} \middle| {x} \right)}}{2}\right)}{x} + \frac{2 {G_{4, 4}^{3, 1}\left(\begin{matrix} 0 & \frac{5}{2}, \frac{5}{2}, \frac{5}{2} \\\frac{3}{2}, \frac{3}{2}, \frac{3}{2} & 0 \end{matrix} \middle| {x} \right)}}{x} & \text{otherwise} \end{cases}\, dx
Numerical answer [src]
0.592592592592593
0.592592592592593

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