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

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

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The solution

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  3           
  /           
 |            
 |  2*x - 1   
 |  ------- dx
 |   x + 3    
 |            
/             
2
∫232x−1x+3 dx\int\limits_{2}^{3} \frac{2 x - 1}{x + 3}\, dx 
Integral((2*x - 1)/(x + 3), (x, 2, 3))
Detail solution

  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 dudu:

      ∫u−1u+6 du\int \frac{u - 1}{u + 6}\, du

      1. Let u=u+6u = u + 6.

        Then let du=dudu = du and substitute dudu:

        ∫u−7u du\int \frac{u - 7}{u}\, du

        1. Rewrite the integrand:

          u−7u=1−7u\frac{u - 7}{u} = 1 - \frac{7}{u}

        2. Integrate term-by-term:

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

            ∫1 du=u\int 1\, du = u

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

            ∫(−7u) du=−7∫1u du\int \left(- \frac{7}{u}\right)\, du = - 7 \int \frac{1}{u}\, du

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

            So, the result is: −7log⁡(u)- 7 \log{\left(u \right)}

          The result is: u−7log⁡(u)u - 7 \log{\left(u \right)}

        Now substitute uu back in:

        u−7log⁡(u+6)+6u - 7 \log{\left(u + 6 \right)} + 6

      Now substitute uu back in:

      2x−7log⁡(2x+6)+62 x - 7 \log{\left(2 x + 6 \right)} + 6

    Method #2

    1. Rewrite the integrand:

      2x−1x+3=2−7x+3\frac{2 x - 1}{x + 3} = 2 - \frac{7}{x + 3}

    2. Integrate term-by-term:

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

        ∫2 dx=2x\int 2\, dx = 2 x

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

        ∫(−7x+3) dx=−7∫1x+3 dx\int \left(- \frac{7}{x + 3}\right)\, dx = - 7 \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: −7log⁡(x+3)- 7 \log{\left(x + 3 \right)}

      The result is: 2x−7log⁡(x+3)2 x - 7 \log{\left(x + 3 \right)}

    Method #3

    1. Rewrite the integrand:

      2x−1x+3=2xx+3−1x+3\frac{2 x - 1}{x + 3} = \frac{2 x}{x + 3} - \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:

        ∫2xx+3 dx=2∫xx+3 dx\int \frac{2 x}{x + 3}\, dx = 2 \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: 2x−6log⁡(x+3)2 x - 6 \log{\left(x + 3 \right)}

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

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

      The result is: 2x−6log⁡(x+3)−log⁡(x+3)2 x - 6 \log{\left(x + 3 \right)} - \log{\left(x + 3 \right)}

  2. Add the constant of integration:

    2x−7log⁡(2x+6)+6+constant2 x - 7 \log{\left(2 x + 6 \right)} + 6+ \mathrm{constant}

The answer is:

2x−7log⁡(2x+6)+6+constant2 x - 7 \log{\left(2 x + 6 \right)} + 6+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                         
 |                                          
 | 2*x - 1                                  
 | ------- dx = 6 + C - 7*log(6 + 2*x) + 2*x
 |  x + 3                                   
 |                                          
/
∫2x−1x+3 dx=C+2x−7log⁡(2x+6)+6\int \frac{2 x - 1}{x + 3}\, dx = C + 2 x - 7 \log{\left(2 x + 6 \right)} + 6
Simplify
The graph
2.003.002.102.202.302.402.502.602.702.802.90-1010
Plot the graph f(x)
The answer [src] 
2 - 7*log(6) + 7*log(5)
−7log⁡(6)+2+7log⁡(5)- 7 \log{\left(6 \right)} + 2 + 7 \log{\left(5 \right)} 
=
= 
2 - 7*log(6) + 7*log(5)
−7log⁡(6)+2+7log⁡(5)- 7 \log{\left(6 \right)} + 2 + 7 \log{\left(5 \right)} 
2 - 7*log(6) + 7*log(5)
Numerical answer [src] 
0.723749102442318
0.723749102442318

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

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