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Solving integrals/ ((x^5-2x)*ln3x)/x

Integral of ((x^5-2x)*ln3x)/x dx

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∫31(x5−2x)log⁡(3x)x dx\int\limits_{3}^{1} \frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x}\, dx 
Integral(((x^5 - 2*x)*log(3*x))/x, (x, 3, 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 dudu:

      ∫2u4log⁡(1u)+2u4log⁡(3)−log⁡(1u)−log⁡(3)u6 du\int \frac{2 u^{4} \log{\left(\frac{1}{u} \right)} + 2 u^{4} \log{\left(3 \right)} - \log{\left(\frac{1}{u} \right)} - \log{\left(3 \right)}}{u^{6}}\, du

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

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

        ∫(ue5u−2ueu+e5ulog⁡(3)−2eulog⁡(3)) du\int \left(u e^{5 u} - 2 u e^{u} + e^{5 u} \log{\left(3 \right)} - 2 e^{u} \log{\left(3 \right)}\right)\, du

        1. Integrate term-by-term:

          1. Use integration by parts:

            ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\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)=e5u\operatorname{dv}{\left(u \right)} = e^{5 u}.

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

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

            1. Let u=5uu = 5 u.

              Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

              ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

                So, the result is: eu5\frac{e^{u}}{5}

              Now substitute uu back in:

              e5u5\frac{e^{5 u}}{5}

            Now evaluate the sub-integral.

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

            ∫e5u5 du=∫e5u du5\int \frac{e^{5 u}}{5}\, du = \frac{\int e^{5 u}\, du}{5}

            1. Let u=5uu = 5 u.

              Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

              ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

                So, the result is: eu5\frac{e^{u}}{5}

              Now substitute uu back in:

              e5u5\frac{e^{5 u}}{5}

            So, the result is: e5u25\frac{e^{5 u}}{25}

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

            ∫(−2ueu) du=−2∫ueu du\int \left(- 2 u e^{u}\right)\, du = - 2 \int u e^{u}\, du

            1. Use integration by parts:

              ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\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)=eu\operatorname{dv}{\left(u \right)} = e^{u}.

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

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

              1. The integral of the exponential function is itself.

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

              Now evaluate the sub-integral.

            2. The integral of the exponential function is itself.

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

            So, the result is: −2ueu+2eu- 2 u e^{u} + 2 e^{u}

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

            ∫e5ulog⁡(3) du=log⁡(3)∫e5u du\int e^{5 u} \log{\left(3 \right)}\, du = \log{\left(3 \right)} \int e^{5 u}\, du

            1. Let u=5uu = 5 u.

              Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

              ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

                So, the result is: eu5\frac{e^{u}}{5}

              Now substitute uu back in:

              e5u5\frac{e^{5 u}}{5}

            So, the result is: e5ulog⁡(3)5\frac{e^{5 u} \log{\left(3 \right)}}{5}

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

            ∫(−2eulog⁡(3)) du=−2log⁡(3)∫eu du\int \left(- 2 e^{u} \log{\left(3 \right)}\right)\, du = - 2 \log{\left(3 \right)} \int e^{u}\, du

            1. The integral of the exponential function is itself.

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

            So, the result is: −2eulog⁡(3)- 2 e^{u} \log{\left(3 \right)}

          The result is: ue5u5−2ueu−e5u25+e5ulog⁡(3)5−2eulog⁡(3)+2eu\frac{u e^{5 u}}{5} - 2 u e^{u} - \frac{e^{5 u}}{25} + \frac{e^{5 u} \log{\left(3 \right)}}{5} - 2 e^{u} \log{\left(3 \right)} + 2 e^{u}

        Now substitute uu back in:

        −2log⁡(1u)u−2log⁡(3)u+2u+log⁡(1u)5u5−125u5+log⁡(3)5u5- \frac{2 \log{\left(\frac{1}{u} \right)}}{u} - \frac{2 \log{\left(3 \right)}}{u} + \frac{2}{u} + \frac{\log{\left(\frac{1}{u} \right)}}{5 u^{5}} - \frac{1}{25 u^{5}} + \frac{\log{\left(3 \right)}}{5 u^{5}}

      Now substitute uu back in:

      x5log⁡(x)5−x525+x5log⁡(3)5−2xlog⁡(x)−2xlog⁡(3)+2x\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x

    Method #2

    1. Rewrite the integrand:

      (x5−2x)log⁡(3x)x=x4log⁡(x)+x4log⁡(3)−2log⁡(x)−2log⁡(3)\frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x} = x^{4} \log{\left(x \right)} + x^{4} \log{\left(3 \right)} - 2 \log{\left(x \right)} - 2 \log{\left(3 \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:

        ∫ue5u du\int u e^{5 u}\, du

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\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)=e5u\operatorname{dv}{\left(u \right)} = e^{5 u}.

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

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

          1. Let u=5uu = 5 u.

            Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

            ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5u5\frac{e^{5 u}}{5}

          Now evaluate the sub-integral.

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

          ∫e5u5 du=∫e5u du5\int \frac{e^{5 u}}{5}\, du = \frac{\int e^{5 u}\, du}{5}

          1. Let u=5uu = 5 u.

            Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

            ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5u5\frac{e^{5 u}}{5}

          So, the result is: e5u25\frac{e^{5 u}}{25}

        Now substitute uu back in:

        x5log⁡(x)5−x525\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25}

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

        ∫x4log⁡(3) dx=log⁡(3)∫x4 dx\int x^{4} \log{\left(3 \right)}\, dx = \log{\left(3 \right)} \int x^{4}\, dx

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

          ∫x4 dx=x55\int x^{4}\, dx = \frac{x^{5}}{5}

        So, the result is: x5log⁡(3)5\frac{x^{5} \log{\left(3 \right)}}{5}

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

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

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\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)=1\operatorname{dv}{\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. The integral of a constant is the constant times the variable of integration:

            ∫1 dx=x\int 1\, dx = x

          Now evaluate the sub-integral.

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

          ∫1 dx=x\int 1\, dx = x

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

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

        ∫(−2log⁡(3)) dx=−2xlog⁡(3)\int \left(- 2 \log{\left(3 \right)}\right)\, dx = - 2 x \log{\left(3 \right)}

      The result is: x5log⁡(x)5−x525+x5log⁡(3)5−2xlog⁡(x)−2xlog⁡(3)+2x\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x

    Method #3

    1. Rewrite the integrand:

      (x5−2x)log⁡(3x)x=x4log⁡(x)+x4log⁡(3)−2log⁡(x)−2log⁡(3)\frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x} = x^{4} \log{\left(x \right)} + x^{4} \log{\left(3 \right)} - 2 \log{\left(x \right)} - 2 \log{\left(3 \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:

        ∫ue5u du\int u e^{5 u}\, du

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\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)=e5u\operatorname{dv}{\left(u \right)} = e^{5 u}.

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

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

          1. Let u=5uu = 5 u.

            Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

            ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5u5\frac{e^{5 u}}{5}

          Now evaluate the sub-integral.

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

          ∫e5u5 du=∫e5u du5\int \frac{e^{5 u}}{5}\, du = \frac{\int e^{5 u}\, du}{5}

          1. Let u=5uu = 5 u.

            Then let du=5dudu = 5 du and substitute du5\frac{du}{5}:

            ∫eu5 du\int \frac{e^{u}}{5}\, 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.

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

              So, the result is: eu5\frac{e^{u}}{5}

            Now substitute uu back in:

            e5u5\frac{e^{5 u}}{5}

          So, the result is: e5u25\frac{e^{5 u}}{25}

        Now substitute uu back in:

        x5log⁡(x)5−x525\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25}

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

        ∫x4log⁡(3) dx=log⁡(3)∫x4 dx\int x^{4} \log{\left(3 \right)}\, dx = \log{\left(3 \right)} \int x^{4}\, dx

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

          ∫x4 dx=x55\int x^{4}\, dx = \frac{x^{5}}{5}

        So, the result is: x5log⁡(3)5\frac{x^{5} \log{\left(3 \right)}}{5}

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

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

        1. Use integration by parts:

          ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\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)=1\operatorname{dv}{\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. The integral of a constant is the constant times the variable of integration:

            ∫1 dx=x\int 1\, dx = x

          Now evaluate the sub-integral.

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

          ∫1 dx=x\int 1\, dx = x

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

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

        ∫(−2log⁡(3)) dx=−2xlog⁡(3)\int \left(- 2 \log{\left(3 \right)}\right)\, dx = - 2 x \log{\left(3 \right)}

      The result is: x5log⁡(x)5−x525+x5log⁡(3)5−2xlog⁡(x)−2xlog⁡(3)+2x\frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x

  2. Now simplify:

    x(5x4log⁡(x)−x4+x4log⁡(243)−50log⁡(x)−log⁡(717897987691852588770249)+50)25\frac{x \left(5 x^{4} \log{\left(x \right)} - x^{4} + x^{4} \log{\left(243 \right)} - 50 \log{\left(x \right)} - \log{\left(717897987691852588770249 \right)} + 50\right)}{25}

  3. Add the constant of integration:

    x(5x4log⁡(x)−x4+x4log⁡(243)−50log⁡(x)−log⁡(717897987691852588770249)+50)25+constant\frac{x \left(5 x^{4} \log{\left(x \right)} - x^{4} + x^{4} \log{\left(243 \right)} - 50 \log{\left(x \right)} - \log{\left(717897987691852588770249 \right)} + 50\right)}{25}+ \mathrm{constant}

The answer is:

x(5x4log⁡(x)−x4+x4log⁡(243)−50log⁡(x)−log⁡(717897987691852588770249)+50)25+constant\frac{x \left(5 x^{4} \log{\left(x \right)} - x^{4} + x^{4} \log{\left(243 \right)} - 50 \log{\left(x \right)} - \log{\left(717897987691852588770249 \right)} + 50\right)}{25}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                       
 |                                                                                        
 | / 5      \                          5                              5           5       
 | \x  - 2*x/*log(3*x)                x                              x *log(3)   x *log(x)
 | ------------------- dx = C + 2*x - -- - 2*x*log(3) - 2*x*log(x) + --------- + ---------
 |          x                         25                                 5           5    
 |                                                                                        
/
∫(x5−2x)log⁡(3x)x dx=C+x5log⁡(x)5−x525+x5log⁡(3)5−2xlog⁡(x)−2xlog⁡(3)+2x\int \frac{\left(x^{5} - 2 x\right) \log{\left(3 x \right)}}{x}\, dx = C + \frac{x^{5} \log{\left(x \right)}}{5} - \frac{x^{5}}{25} + \frac{x^{5} \log{\left(3 \right)}}{5} - 2 x \log{\left(x \right)} - 2 x \log{\left(3 \right)} + 2 x
Simplify
The graph
1.03.01.21.41.61.82.02.22.42.62.8-200200
Plot the graph f(x)
The answer [src] 
142   213*log(9)   9*log(3)
--- - ---------- - --------
 25       5           5
−213log⁡(9)5−9log⁡(3)5+14225- \frac{213 \log{\left(9 \right)}}{5} - \frac{9 \log{\left(3 \right)}}{5} + \frac{142}{25} 
=
= 
142   213*log(9)   9*log(3)
--- - ---------- - --------
 25       5           5
−213log⁡(9)5−9log⁡(3)5+14225- \frac{213 \log{\left(9 \right)}}{5} - \frac{9 \log{\left(3 \right)}}{5} + \frac{142}{25} 
142/25 - 213*log(9)/5 - 9*log(3)/5
Numerical answer [src] 
-89.8992691141255
-89.8992691141255

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

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