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Solving integrals/ (2x-1)/(x+3)(x^2-4)

Integral of (2x-1)/(x+3)(x^2-4) dx

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The solution

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  1                    
  /                    
 |                     
 |  2*x - 1 / 2    \   
 |  -------*\x  - 4/ dx
 |   x + 3             
 |                     
/                      
0
∫012x−1x+3(x2−4) dx\int\limits_{0}^{1} \frac{2 x - 1}{x + 3} \left(x^{2} - 4\right)\, dx 
Integral(((2*x - 1)/(x + 3))*(x^2 - 4), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      2x−1x+3(x2−4)=2x2−7x+13−35x+3\frac{2 x - 1}{x + 3} \left(x^{2} - 4\right) = 2 x^{2} - 7 x + 13 - \frac{35}{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:

        ∫2x2 dx=2∫x2 dx\int 2 x^{2}\, dx = 2 \int x^{2}\, dx

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

          ∫x2 dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

        So, the result is: 2x33\frac{2 x^{3}}{3}

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

        ∫(−7x) dx=−7∫x dx\int \left(- 7 x\right)\, dx = - 7 \int x\, dx

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

          ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

        So, the result is: −7x22- \frac{7 x^{2}}{2}

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

        ∫13 dx=13x\int 13\, dx = 13 x

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

        ∫(−35x+3) dx=−35∫1x+3 dx\int \left(- \frac{35}{x + 3}\right)\, dx = - 35 \int \frac{1}{x + 3}\, dx

        1. Let u=x+3u = x + 3.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\int \frac{1}{u}\, du

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

          Now substitute uu back in:

          log⁡(x+3)\log{\left(x + 3 \right)}

        So, the result is: −35log⁡(x+3)- 35 \log{\left(x + 3 \right)}

      The result is: 2x33−7x22+13x−35log⁡(x+3)\frac{2 x^{3}}{3} - \frac{7 x^{2}}{2} + 13 x - 35 \log{\left(x + 3 \right)}

    Method #2

    1. Rewrite the integrand:

      2x−1x+3(x2−4)=2x3−x2−8x+4x+3\frac{2 x - 1}{x + 3} \left(x^{2} - 4\right) = \frac{2 x^{3} - x^{2} - 8 x + 4}{x + 3}

    2. Rewrite the integrand:

      2x3−x2−8x+4x+3=2x2−7x+13−35x+3\frac{2 x^{3} - x^{2} - 8 x + 4}{x + 3} = 2 x^{2} - 7 x + 13 - \frac{35}{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:

        ∫2x2 dx=2∫x2 dx\int 2 x^{2}\, dx = 2 \int x^{2}\, dx

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

          ∫x2 dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

        So, the result is: 2x33\frac{2 x^{3}}{3}

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

        ∫(−7x) dx=−7∫x dx\int \left(- 7 x\right)\, dx = - 7 \int x\, dx

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

          ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

        So, the result is: −7x22- \frac{7 x^{2}}{2}

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

        ∫13 dx=13x\int 13\, dx = 13 x

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

        ∫(−35x+3) dx=−35∫1x+3 dx\int \left(- \frac{35}{x + 3}\right)\, dx = - 35 \int \frac{1}{x + 3}\, dx

        1. Let u=x+3u = x + 3.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\int \frac{1}{u}\, du

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

          Now substitute uu back in:

          log⁡(x+3)\log{\left(x + 3 \right)}

        So, the result is: −35log⁡(x+3)- 35 \log{\left(x + 3 \right)}

      The result is: 2x33−7x22+13x−35log⁡(x+3)\frac{2 x^{3}}{3} - \frac{7 x^{2}}{2} + 13 x - 35 \log{\left(x + 3 \right)}

    Method #3

    1. Rewrite the integrand:

      2x−1x+3(x2−4)=2x3x+3−x2x+3−8xx+3+4x+3\frac{2 x - 1}{x + 3} \left(x^{2} - 4\right) = \frac{2 x^{3}}{x + 3} - \frac{x^{2}}{x + 3} - \frac{8 x}{x + 3} + \frac{4}{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:

        ∫2x3x+3 dx=2∫x3x+3 dx\int \frac{2 x^{3}}{x + 3}\, dx = 2 \int \frac{x^{3}}{x + 3}\, dx

        1. Rewrite the integrand:

          x3x+3=x2−3x+9−27x+3\frac{x^{3}}{x + 3} = x^{2} - 3 x + 9 - \frac{27}{x + 3}

        2. Integrate term-by-term:

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

            ∫x2 dx=x33\int x^{2}\, dx = \frac{x^{3}}{3}

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

            ∫(−3x) dx=−3∫x dx\int \left(- 3 x\right)\, dx = - 3 \int x\, dx

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

              ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

            So, the result is: −3x22- \frac{3 x^{2}}{2}

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

            ∫9 dx=9x\int 9\, dx = 9 x

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

            ∫(−27x+3) dx=−27∫1x+3 dx\int \left(- \frac{27}{x + 3}\right)\, dx = - 27 \int \frac{1}{x + 3}\, dx

            1. Let u=x+3u = x + 3.

              Then let du=dxdu = dx and substitute dudu:

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

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

              Now substitute uu back in:

              log⁡(x+3)\log{\left(x + 3 \right)}

            So, the result is: −27log⁡(x+3)- 27 \log{\left(x + 3 \right)}

          The result is: x33−3x22+9x−27log⁡(x+3)\frac{x^{3}}{3} - \frac{3 x^{2}}{2} + 9 x - 27 \log{\left(x + 3 \right)}

        So, the result is: 2x33−3x2+18x−54log⁡(x+3)\frac{2 x^{3}}{3} - 3 x^{2} + 18 x - 54 \log{\left(x + 3 \right)}

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

        ∫(−x2x+3) dx=−∫x2x+3 dx\int \left(- \frac{x^{2}}{x + 3}\right)\, dx = - \int \frac{x^{2}}{x + 3}\, dx

        1. Rewrite the integrand:

          x2x+3=x−3+9x+3\frac{x^{2}}{x + 3} = x - 3 + \frac{9}{x + 3}

        2. Integrate term-by-term:

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

            ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

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

            ∫(−3) dx=−3x\int \left(-3\right)\, dx = - 3 x

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

            ∫9x+3 dx=9∫1x+3 dx\int \frac{9}{x + 3}\, dx = 9 \int \frac{1}{x + 3}\, dx

            1. Let u=x+3u = x + 3.

              Then let du=dxdu = dx and substitute dudu:

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

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

              Now substitute uu back in:

              log⁡(x+3)\log{\left(x + 3 \right)}

            So, the result is: 9log⁡(x+3)9 \log{\left(x + 3 \right)}

          The result is: x22−3x+9log⁡(x+3)\frac{x^{2}}{2} - 3 x + 9 \log{\left(x + 3 \right)}

        So, the result is: −x22+3x−9log⁡(x+3)- \frac{x^{2}}{2} + 3 x - 9 \log{\left(x + 3 \right)}

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

        ∫(−8xx+3) dx=−8∫xx+3 dx\int \left(- \frac{8 x}{x + 3}\right)\, dx = - 8 \int \frac{x}{x + 3}\, dx

        1. Rewrite the integrand:

          xx+3=1−3x+3\frac{x}{x + 3} = 1 - \frac{3}{x + 3}

        2. Integrate term-by-term:

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

            ∫1 dx=x\int 1\, dx = x

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

            ∫(−3x+3) dx=−3∫1x+3 dx\int \left(- \frac{3}{x + 3}\right)\, dx = - 3 \int \frac{1}{x + 3}\, dx

            1. Let u=x+3u = x + 3.

              Then let du=dxdu = dx and substitute dudu:

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

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

              Now substitute uu back in:

              log⁡(x+3)\log{\left(x + 3 \right)}

            So, the result is: −3log⁡(x+3)- 3 \log{\left(x + 3 \right)}

          The result is: x−3log⁡(x+3)x - 3 \log{\left(x + 3 \right)}

        So, the result is: −8x+24log⁡(x+3)- 8 x + 24 \log{\left(x + 3 \right)}

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

        ∫4x+3 dx=4∫1x+3 dx\int \frac{4}{x + 3}\, dx = 4 \int \frac{1}{x + 3}\, dx

        1. Let u=x+3u = x + 3.

          Then let du=dxdu = dx and substitute dudu:

          ∫1u du\int \frac{1}{u}\, du

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

          Now substitute uu back in:

          log⁡(x+3)\log{\left(x + 3 \right)}

        So, the result is: 4log⁡(x+3)4 \log{\left(x + 3 \right)}

      The result is: 2x33−7x22+13x+4log⁡(x+3)−39log⁡(x+3)\frac{2 x^{3}}{3} - \frac{7 x^{2}}{2} + 13 x + 4 \log{\left(x + 3 \right)} - 39 \log{\left(x + 3 \right)}

  2. Add the constant of integration:

    2x33−7x22+13x−35log⁡(x+3)+constant\frac{2 x^{3}}{3} - \frac{7 x^{2}}{2} + 13 x - 35 \log{\left(x + 3 \right)}+ \mathrm{constant}

The answer is:

2x33−7x22+13x−35log⁡(x+3)+constant\frac{2 x^{3}}{3} - \frac{7 x^{2}}{2} + 13 x - 35 \log{\left(x + 3 \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                            
 |                                                     2      3
 | 2*x - 1 / 2    \                                 7*x    2*x 
 | -------*\x  - 4/ dx = C - 35*log(3 + x) + 13*x - ---- + ----
 |  x + 3                                            2      3  
 |                                                             
/
∫2x−1x+3(x2−4) dx=C+2x33−7x22+13x−35log⁡(x+3)\int \frac{2 x - 1}{x + 3} \left(x^{2} - 4\right)\, dx = C + \frac{2 x^{3}}{3} - \frac{7 x^{2}}{2} + 13 x - 35 \log{\left(x + 3 \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90-5050
Plot the graph f(x)
The answer [src] 
61/6 - 35*log(4) + 35*log(3)
−35log⁡(4)+616+35log⁡(3)- 35 \log{\left(4 \right)} + \frac{61}{6} + 35 \log{\left(3 \right)} 
=
= 
61/6 - 35*log(4) + 35*log(3)
−35log⁡(4)+616+35log⁡(3)- 35 \log{\left(4 \right)} + \frac{61}{6} + 35 \log{\left(3 \right)} 
61/6 - 35*log(4) + 35*log(3)
Numerical answer [src] 
0.0977941308543342
0.0977941308543342

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

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