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Solving integrals/ arctg2x-(3/(x^4))arctg2x

Integral of arctg2x-(3/(x^4))arctg2x dx

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∫01(−3x4atan⁡(2x)+atan⁡(2x)) dx\int\limits_{0}^{1} \left(- \frac{3}{x^{4}} \operatorname{atan}{\left(2 x \right)} + \operatorname{atan}{\left(2 x \right)}\right)\, dx 
Integral(atan(2*x) - 3/x^4*atan(2*x), (x, 0, 1))
Detail solution

  1. Integrate term-by-term:

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

      ∫(−3x4atan⁡(2x)) dx=−∫3atan⁡(2x)x4 dx\int \left(- \frac{3}{x^{4}} \operatorname{atan}{\left(2 x \right)}\right)\, dx = - \int \frac{3 \operatorname{atan}{\left(2 x \right)}}{x^{4}}\, dx

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

        ∫3atan⁡(2x)x4 dx=3∫atan⁡(2x)x4 dx\int \frac{3 \operatorname{atan}{\left(2 x \right)}}{x^{4}}\, dx = 3 \int \frac{\operatorname{atan}{\left(2 x \right)}}{x^{4}}\, 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)=atan⁡(2x)u{\left(x \right)} = \operatorname{atan}{\left(2 x \right)} and let dv⁡(x)=1x4\operatorname{dv}{\left(x \right)} = \frac{1}{x^{4}}.

          Then du⁡(x)=24x2+1\operatorname{du}{\left(x \right)} = \frac{2}{4 x^{2} + 1}.

          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 n≠−1n \neq -1:

            ∫1x4 dx=−13x3\int \frac{1}{x^{4}}\, dx = - \frac{1}{3 x^{3}}

          Now evaluate the sub-integral.

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

          ∫(−23x3(4x2+1)) dx=−2∫1x3(4x2+1) dx3\int \left(- \frac{2}{3 x^{3} \left(4 x^{2} + 1\right)}\right)\, dx = - \frac{2 \int \frac{1}{x^{3} \left(4 x^{2} + 1\right)}\, dx}{3}

          1. There are multiple ways to do this integral.

            Method #1

            1. Let u=x2u = x^{2}.

              Then let du=2xdxdu = 2 x dx and substitute dudu:

              ∫18u3+2u2 du\int \frac{1}{8 u^{3} + 2 u^{2}}\, du

              1. Rewrite the integrand:

                18u3+2u2=84u+1−2u+12u2\frac{1}{8 u^{3} + 2 u^{2}} = \frac{8}{4 u + 1} - \frac{2}{u} + \frac{1}{2 u^{2}}

              2. Integrate term-by-term:

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

                  ∫84u+1 du=8∫14u+1 du\int \frac{8}{4 u + 1}\, du = 8 \int \frac{1}{4 u + 1}\, du

                  1. Let u=4u+1u = 4 u + 1.

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

                    ∫14u du\int \frac{1}{4 u}\, du

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

                      ∫1u du=∫1u du4\int \frac{1}{u}\, du = \frac{\int \frac{1}{u}\, du}{4}

                      1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

                      So, the result is: log⁡(u)4\frac{\log{\left(u \right)}}{4}

                    Now substitute uu back in:

                    log⁡(4u+1)4\frac{\log{\left(4 u + 1 \right)}}{4}

                  So, the result is: 2log⁡(4u+1)2 \log{\left(4 u + 1 \right)}

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

                  ∫(−2u) du=−2∫1u du\int \left(- \frac{2}{u}\right)\, du = - 2 \int \frac{1}{u}\, du

                  1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

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

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

                  ∫12u2 du=∫1u2 du2\int \frac{1}{2 u^{2}}\, du = \frac{\int \frac{1}{u^{2}}\, du}{2}

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

                    ∫1u2 du=−1u\int \frac{1}{u^{2}}\, du = - \frac{1}{u}

                  So, the result is: −12u- \frac{1}{2 u}

                The result is: −2log⁡(u)+2log⁡(4u+1)−12u- 2 \log{\left(u \right)} + 2 \log{\left(4 u + 1 \right)} - \frac{1}{2 u}

              Now substitute uu back in:

              −2log⁡(x2)+2log⁡(4x2+1)−12x2- 2 \log{\left(x^{2} \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}

            Method #2

            1. Rewrite the integrand:

              1x3(4x2+1)=16x4x2+1−4x+1x3\frac{1}{x^{3} \left(4 x^{2} + 1\right)} = \frac{16 x}{4 x^{2} + 1} - \frac{4}{x} + \frac{1}{x^{3}}

            2. Integrate term-by-term:

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

                ∫16x4x2+1 dx=16∫x4x2+1 dx\int \frac{16 x}{4 x^{2} + 1}\, dx = 16 \int \frac{x}{4 x^{2} + 1}\, dx

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

                  ∫x4x2+1 dx=∫8x4x2+1 dx8\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}

                  1. Let u=4x2+1u = 4 x^{2} + 1.

                    Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

                    ∫18u du\int \frac{1}{8 u}\, du

                    1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

                    Now substitute uu back in:

                    log⁡(4x2+1)\log{\left(4 x^{2} + 1 \right)}

                  So, the result is: log⁡(4x2+1)8\frac{\log{\left(4 x^{2} + 1 \right)}}{8}

                So, the result is: 2log⁡(4x2+1)2 \log{\left(4 x^{2} + 1 \right)}

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

                ∫(−4x) dx=−4∫1x dx\int \left(- \frac{4}{x}\right)\, dx = - 4 \int \frac{1}{x}\, dx

                1. The integral of 1x\frac{1}{x} is log⁡(x)\log{\left(x \right)}.

                So, the result is: −4log⁡(x)- 4 \log{\left(x \right)}

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

                ∫1x3 dx=−12x2\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}

              The result is: −4log⁡(x)+2log⁡(4x2+1)−12x2- 4 \log{\left(x \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}

            Method #3

            1. Rewrite the integrand:

              1x3(4x2+1)=14x5+x3\frac{1}{x^{3} \left(4 x^{2} + 1\right)} = \frac{1}{4 x^{5} + x^{3}}

            2. Rewrite the integrand:

              14x5+x3=16x4x2+1−4x+1x3\frac{1}{4 x^{5} + x^{3}} = \frac{16 x}{4 x^{2} + 1} - \frac{4}{x} + \frac{1}{x^{3}}

            3. Integrate term-by-term:

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

                ∫16x4x2+1 dx=16∫x4x2+1 dx\int \frac{16 x}{4 x^{2} + 1}\, dx = 16 \int \frac{x}{4 x^{2} + 1}\, dx

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

                  ∫x4x2+1 dx=∫8x4x2+1 dx8\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}

                  1. Let u=4x2+1u = 4 x^{2} + 1.

                    Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

                    ∫18u du\int \frac{1}{8 u}\, du

                    1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

                    Now substitute uu back in:

                    log⁡(4x2+1)\log{\left(4 x^{2} + 1 \right)}

                  So, the result is: log⁡(4x2+1)8\frac{\log{\left(4 x^{2} + 1 \right)}}{8}

                So, the result is: 2log⁡(4x2+1)2 \log{\left(4 x^{2} + 1 \right)}

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

                ∫(−4x) dx=−4∫1x dx\int \left(- \frac{4}{x}\right)\, dx = - 4 \int \frac{1}{x}\, dx

                1. The integral of 1x\frac{1}{x} is log⁡(x)\log{\left(x \right)}.

                So, the result is: −4log⁡(x)- 4 \log{\left(x \right)}

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

                ∫1x3 dx=−12x2\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}

              The result is: −4log⁡(x)+2log⁡(4x2+1)−12x2- 4 \log{\left(x \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}

            Method #4

            1. Rewrite the integrand:

              1x3(4x2+1)=14x5+x3\frac{1}{x^{3} \left(4 x^{2} + 1\right)} = \frac{1}{4 x^{5} + x^{3}}

            2. Rewrite the integrand:

              14x5+x3=16x4x2+1−4x+1x3\frac{1}{4 x^{5} + x^{3}} = \frac{16 x}{4 x^{2} + 1} - \frac{4}{x} + \frac{1}{x^{3}}

            3. Integrate term-by-term:

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

                ∫16x4x2+1 dx=16∫x4x2+1 dx\int \frac{16 x}{4 x^{2} + 1}\, dx = 16 \int \frac{x}{4 x^{2} + 1}\, dx

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

                  ∫x4x2+1 dx=∫8x4x2+1 dx8\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}

                  1. Let u=4x2+1u = 4 x^{2} + 1.

                    Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

                    ∫18u du\int \frac{1}{8 u}\, du

                    1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

                    Now substitute uu back in:

                    log⁡(4x2+1)\log{\left(4 x^{2} + 1 \right)}

                  So, the result is: log⁡(4x2+1)8\frac{\log{\left(4 x^{2} + 1 \right)}}{8}

                So, the result is: 2log⁡(4x2+1)2 \log{\left(4 x^{2} + 1 \right)}

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

                ∫(−4x) dx=−4∫1x dx\int \left(- \frac{4}{x}\right)\, dx = - 4 \int \frac{1}{x}\, dx

                1. The integral of 1x\frac{1}{x} is log⁡(x)\log{\left(x \right)}.

                So, the result is: −4log⁡(x)- 4 \log{\left(x \right)}

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

                ∫1x3 dx=−12x2\int \frac{1}{x^{3}}\, dx = - \frac{1}{2 x^{2}}

              The result is: −4log⁡(x)+2log⁡(4x2+1)−12x2- 4 \log{\left(x \right)} + 2 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{2 x^{2}}

          So, the result is: 4log⁡(x2)3−4log⁡(4x2+1)3+13x2\frac{4 \log{\left(x^{2} \right)}}{3} - \frac{4 \log{\left(4 x^{2} + 1 \right)}}{3} + \frac{1}{3 x^{2}}

        So, the result is: −4log⁡(x2)+4log⁡(4x2+1)−1x2−atan⁡(2x)x3- 4 \log{\left(x^{2} \right)} + 4 \log{\left(4 x^{2} + 1 \right)} - \frac{1}{x^{2}} - \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}

      So, the result is: 4log⁡(x2)−4log⁡(4x2+1)+1x2+atan⁡(2x)x34 \log{\left(x^{2} \right)} - 4 \log{\left(4 x^{2} + 1 \right)} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}

    1. There are multiple ways to do this integral.

      Method #1

      1. Let u=2xu = 2 x.

        Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

        ∫atan⁡(u)2 du\int \frac{\operatorname{atan}{\left(u \right)}}{2}\, du

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

          ∫atan⁡(u) du=∫atan⁡(u) du2\int \operatorname{atan}{\left(u \right)}\, du = \frac{\int \operatorname{atan}{\left(u \right)}\, du}{2}

          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)=atan⁡(u)u{\left(u \right)} = \operatorname{atan}{\left(u \right)} and let dv⁡(u)=1\operatorname{dv}{\left(u \right)} = 1.

            Then du⁡(u)=1u2+1\operatorname{du}{\left(u \right)} = \frac{1}{u^{2} + 1}.

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

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

              ∫1 du=u\int 1\, du = u

            Now evaluate the sub-integral.

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

            ∫uu2+1 du=∫2uu2+1 du2\int \frac{u}{u^{2} + 1}\, du = \frac{\int \frac{2 u}{u^{2} + 1}\, du}{2}

            1. Let u=u2+1u = u^{2} + 1.

              Then let du=2ududu = 2 u du and substitute du2\frac{du}{2}:

              ∫12u du\int \frac{1}{2 u}\, du

              1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

              Now substitute uu back in:

              log⁡(u2+1)\log{\left(u^{2} + 1 \right)}

            So, the result is: log⁡(u2+1)2\frac{\log{\left(u^{2} + 1 \right)}}{2}

          So, the result is: uatan⁡(u)2−log⁡(u2+1)4\frac{u \operatorname{atan}{\left(u \right)}}{2} - \frac{\log{\left(u^{2} + 1 \right)}}{4}

        Now substitute uu back in:

        xatan⁡(2x)−log⁡(4x2+1)4x \operatorname{atan}{\left(2 x \right)} - \frac{\log{\left(4 x^{2} + 1 \right)}}{4}

      Method #2

      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)=atan⁡(2x)u{\left(x \right)} = \operatorname{atan}{\left(2 x \right)} and let dv⁡(x)=1\operatorname{dv}{\left(x \right)} = 1.

        Then du⁡(x)=24x2+1\operatorname{du}{\left(x \right)} = \frac{2}{4 x^{2} + 1}.

        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 times a function is the constant times the integral of the function:

        ∫2x4x2+1 dx=2∫x4x2+1 dx\int \frac{2 x}{4 x^{2} + 1}\, dx = 2 \int \frac{x}{4 x^{2} + 1}\, dx

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

          ∫x4x2+1 dx=∫8x4x2+1 dx8\int \frac{x}{4 x^{2} + 1}\, dx = \frac{\int \frac{8 x}{4 x^{2} + 1}\, dx}{8}

          1. Let u=4x2+1u = 4 x^{2} + 1.

            Then let du=8xdxdu = 8 x dx and substitute du8\frac{du}{8}:

            ∫18u du\int \frac{1}{8 u}\, du

            1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

            Now substitute uu back in:

            log⁡(4x2+1)\log{\left(4 x^{2} + 1 \right)}

          So, the result is: log⁡(4x2+1)8\frac{\log{\left(4 x^{2} + 1 \right)}}{8}

        So, the result is: log⁡(4x2+1)4\frac{\log{\left(4 x^{2} + 1 \right)}}{4}

    The result is: xatan⁡(2x)+4log⁡(x2)−17log⁡(4x2+1)4+1x2+atan⁡(2x)x3x \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}

  2. Add the constant of integration:

    xatan⁡(2x)+4log⁡(x2)−17log⁡(4x2+1)4+1x2+atan⁡(2x)x3+constantx \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}+ \mathrm{constant}

The answer is:

xatan⁡(2x)+4log⁡(x2)−17log⁡(4x2+1)4+1x2+atan⁡(2x)x3+constantx \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                               
 |                                                            /       2\                          
 | /            3           \          1         / 2\   17*log\1 + 4*x /                 atan(2*x)
 | |atan(2*x) - --*atan(2*x)| dx = C + -- + 4*log\x / - ---------------- + x*atan(2*x) + ---------
 | |             4          |           2                      4                              3   
 | \            x           /          x                                                     x    
 |                                                                                                
/
∫(−3x4atan⁡(2x)+atan⁡(2x)) dx=C+xatan⁡(2x)+4log⁡(x2)−17log⁡(4x2+1)4+1x2+atan⁡(2x)x3\int \left(- \frac{3}{x^{4}} \operatorname{atan}{\left(2 x \right)} + \operatorname{atan}{\left(2 x \right)}\right)\, dx = C + x \operatorname{atan}{\left(2 x \right)} + 4 \log{\left(x^{2} \right)} - \frac{17 \log{\left(4 x^{2} + 1 \right)}}{4} + \frac{1}{x^{2}} + \frac{\operatorname{atan}{\left(2 x \right)}}{x^{3}}
Simplify
The answer [src] 
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−∞-\infty 
=
= 
-oo
−∞-\infty 
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Numerical answer [src] 
-5.49219022742095e+38
-5.49219022742095e+38

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

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