Mister Exam

Integral of x³lnxdx dx

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The solution

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01x3log(x)1dx\int\limits_{0}^{1} x^{3} \log{\left(x \right)} 1\, dx
Integral(x^3*log(x)*1, (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:

      ue4udu\int u e^{4 u}\, 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)=uu{\left(u \right)} = u and let dv(u)=e4u\operatorname{dv}{\left(u \right)} = e^{4 u}.

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

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

        1. There are multiple ways to do this integral.

          Method #1

          1. Let u=4uu = 4 u.

            Then let du=4dudu = 4 du and substitute du4\frac{du}{4}:

            eu16du\int \frac{e^{u}}{16}\, du

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

              eu4du=eudu4\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}

              1. The integral of the exponential function is itself.

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

              So, the result is: eu4\frac{e^{u}}{4}

            Now substitute uu back in:

            e4u4\frac{e^{4 u}}{4}

          Method #2

          1. Let u=e4uu = e^{4 u}.

            Then let du=4e4ududu = 4 e^{4 u} du and substitute du4\frac{du}{4}:

            116du\int \frac{1}{16}\, du

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

              14du=1du4\int \frac{1}{4}\, du = \frac{\int 1\, du}{4}

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

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

              So, the result is: u4\frac{u}{4}

            Now substitute uu back in:

            e4u4\frac{e^{4 u}}{4}

        Now evaluate the sub-integral.

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

        e4u4du=e4udu4\int \frac{e^{4 u}}{4}\, du = \frac{\int e^{4 u}\, du}{4}

        1. Let u=4uu = 4 u.

          Then let du=4dudu = 4 du and substitute du4\frac{du}{4}:

          eu16du\int \frac{e^{u}}{16}\, du

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

            eu4du=eudu4\int \frac{e^{u}}{4}\, du = \frac{\int e^{u}\, du}{4}

            1. The integral of the exponential function is itself.

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

            So, the result is: eu4\frac{e^{u}}{4}

          Now substitute uu back in:

          e4u4\frac{e^{4 u}}{4}

        So, the result is: e4u16\frac{e^{4 u}}{16}

      Now substitute uu back in:

      x4log(x)4x416\frac{x^{4} \log{\left(x \right)}}{4} - \frac{x^{4}}{16}

    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)=x3\operatorname{dv}{\left(x \right)} = x^{3}.

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

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

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n1n \neq -1:

        x3dx=x44\int x^{3}\, dx = \frac{x^{4}}{4}

      Now evaluate the sub-integral.

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

      x34dx=x3dx4\int \frac{x^{3}}{4}\, dx = \frac{\int x^{3}\, dx}{4}

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n1n \neq -1:

        x3dx=x44\int x^{3}\, dx = \frac{x^{4}}{4}

      So, the result is: x416\frac{x^{4}}{16}

  2. Now simplify:

    x4(4log(x)1)16\frac{x^{4} \cdot \left(4 \log{\left(x \right)} - 1\right)}{16}

  3. Add the constant of integration:

    x4(4log(x)1)16+constant\frac{x^{4} \cdot \left(4 \log{\left(x \right)} - 1\right)}{16}+ \mathrm{constant}


The answer is:

x4(4log(x)1)16+constant\frac{x^{4} \cdot \left(4 \log{\left(x \right)} - 1\right)}{16}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                   
 |                       4    4       
 |  3                   x    x *log(x)
 | x *log(x)*1 dx = C - -- + ---------
 |                      16       4    
/                                     
x4logx4x416{{x^4\,\log x}\over{4}}-{{x^4}\over{16}}
The answer [src]
-1/16
116-{{1}\over{16}}
=
=
-1/16
116- \frac{1}{16}
Numerical answer [src]
-0.0625
-0.0625

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