Mister Exam

Integral of x^2ln2xdx dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src]
  E               
  /               
 |                
 |   2            
 |  x *log(2*x) dx
 |                
/                 
1                 
1ex2log(2x)dx\int\limits_{1}^{e} x^{2} \log{\left(2 x \right)}\, dx
Integral(x^2*log(2*x), (x, 1, E))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      x2log(2x)=x2log(x)+x2log(2)x^{2} \log{\left(2 x \right)} = x^{2} \log{\left(x \right)} + x^{2} \log{\left(2 \right)}

    2. Integrate term-by-term:

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

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

        ue3udu\int u e^{3 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)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

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

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

          1. Let u=3uu = 3 u.

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

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

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

              False\text{False}

              1. The integral of the exponential function is itself.

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

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

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

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

          e3u3du=e3udu3\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}

          1. Let u=3uu = 3 u.

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

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

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

              False\text{False}

              1. The integral of the exponential function is itself.

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

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

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          So, the result is: e3u9\frac{e^{3 u}}{9}

        Now substitute uu back in:

        x3log(x)3x39\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}

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

        x2log(2)dx=log(2)x2dx\int x^{2} \log{\left(2 \right)}\, dx = \log{\left(2 \right)} \int x^{2}\, dx

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

          x2dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

        So, the result is: x3log(2)3\frac{x^{3} \log{\left(2 \right)}}{3}

      The result is: x3log(x)3x39+x3log(2)3\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} + \frac{x^{3} \log{\left(2 \right)}}{3}

    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(2x)u{\left(x \right)} = \log{\left(2 x \right)} and let dv(x)=x2\operatorname{dv}{\left(x \right)} = x^{2}.

      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:

        x2dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

      Now evaluate the sub-integral.

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

      x23dx=x2dx3\int \frac{x^{2}}{3}\, dx = \frac{\int x^{2}\, dx}{3}

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

        x2dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

      So, the result is: x39\frac{x^{3}}{9}

    Method #3

    1. Rewrite the integrand:

      x2log(2x)=x2log(x)+x2log(2)x^{2} \log{\left(2 x \right)} = x^{2} \log{\left(x \right)} + x^{2} \log{\left(2 \right)}

    2. Integrate term-by-term:

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

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

        ue3udu\int u e^{3 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)=e3u\operatorname{dv}{\left(u \right)} = e^{3 u}.

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

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

          1. Let u=3uu = 3 u.

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

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

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

              False\text{False}

              1. The integral of the exponential function is itself.

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

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

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          Now evaluate the sub-integral.

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

          e3u3du=e3udu3\int \frac{e^{3 u}}{3}\, du = \frac{\int e^{3 u}\, du}{3}

          1. Let u=3uu = 3 u.

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

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

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

              False\text{False}

              1. The integral of the exponential function is itself.

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

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

            Now substitute uu back in:

            e3u3\frac{e^{3 u}}{3}

          So, the result is: e3u9\frac{e^{3 u}}{9}

        Now substitute uu back in:

        x3log(x)3x39\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9}

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

        x2log(2)dx=log(2)x2dx\int x^{2} \log{\left(2 \right)}\, dx = \log{\left(2 \right)} \int x^{2}\, dx

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

          x2dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

        So, the result is: x3log(2)3\frac{x^{3} \log{\left(2 \right)}}{3}

      The result is: x3log(x)3x39+x3log(2)3\frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} + \frac{x^{3} \log{\left(2 \right)}}{3}

  2. Now simplify:

    x3(3log(x)1+log(8))9\frac{x^{3} \left(3 \log{\left(x \right)} - 1 + \log{\left(8 \right)}\right)}{9}

  3. Add the constant of integration:

    x3(3log(x)1+log(8))9+constant\frac{x^{3} \left(3 \log{\left(x \right)} - 1 + \log{\left(8 \right)}\right)}{9}+ \mathrm{constant}


The answer is:

x3(3log(x)1+log(8))9+constant\frac{x^{3} \left(3 \log{\left(x \right)} - 1 + \log{\left(8 \right)}\right)}{9}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                               
 |                       3    3           3       
 |  2                   x    x *log(2)   x *log(x)
 | x *log(2*x) dx = C - -- + --------- + ---------
 |                      9        3           3    
/                                                 
x2log(2x)dx=C+x3log(x)3x39+x3log(2)3\int x^{2} \log{\left(2 x \right)}\, dx = C + \frac{x^{3} \log{\left(x \right)}}{3} - \frac{x^{3}}{9} + \frac{x^{3} \log{\left(2 \right)}}{3}
The graph
1.01.21.41.61.82.02.22.42.6020
The answer [src]
              3    3         
1   log(2)   e    e *log(2*E)
- - ------ - -- + -----------
9     3      9         3     
e39log(2)3+19+e3log(2e)3- \frac{e^{3}}{9} - \frac{\log{\left(2 \right)}}{3} + \frac{1}{9} + \frac{e^{3} \log{\left(2 e \right)}}{3}
=
=
              3    3         
1   log(2)   e    e *log(2*E)
- - ------ - -- + -----------
9     3      9         3     
e39log(2)3+19+e3log(2e)3- \frac{e^{3}}{9} - \frac{\log{\left(2 \right)}}{3} + \frac{1}{9} + \frac{e^{3} \log{\left(2 e \right)}}{3}
1/9 - log(2)/3 - exp(3)/9 + exp(3)*log(2*E)/3
Numerical answer [src]
8.98425912996846
8.98425912996846

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