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(2*x+1)/(x+2)

Integral of (2*x+1)/(x+2) dx

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  1           
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 |  2*x + 1   
 |  ------- dx
 |   x + 2    
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012x+1x+2dx\int\limits_{0}^{1} \frac{2 x + 1}{x + 2}\, dx
Integral((2*x + 1)/(x + 2), (x, 0, 1))
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+4du\int \frac{u + 1}{u + 4}\, du

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

        Then let du=dudu = du and substitute dudu:

        u3udu\int \frac{u - 3}{u}\, du

        1. Rewrite the integrand:

          u3u=13u\frac{u - 3}{u} = 1 - \frac{3}{u}

        2. Integrate term-by-term:

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

            1du=u\int 1\, du = u

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

            (3u)du=31udu\int \left(- \frac{3}{u}\right)\, du = - 3 \int \frac{1}{u}\, du

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

            So, the result is: 3log(u)- 3 \log{\left(u \right)}

          The result is: u3log(u)u - 3 \log{\left(u \right)}

        Now substitute uu back in:

        u3log(u+4)+4u - 3 \log{\left(u + 4 \right)} + 4

      Now substitute uu back in:

      2x3log(2x+4)+42 x - 3 \log{\left(2 x + 4 \right)} + 4

    Method #2

    1. Rewrite the integrand:

      2x+1x+2=23x+2\frac{2 x + 1}{x + 2} = 2 - \frac{3}{x + 2}

    2. Integrate term-by-term:

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

        2dx=2x\int 2\, dx = 2 x

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

        (3x+2)dx=31x+2dx\int \left(- \frac{3}{x + 2}\right)\, dx = - 3 \int \frac{1}{x + 2}\, dx

        1. Let u=x+2u = x + 2.

          Then let du=dxdu = dx and substitute dudu:

          1udu\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+2)\log{\left(x + 2 \right)}

        So, the result is: 3log(x+2)- 3 \log{\left(x + 2 \right)}

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

    Method #3

    1. Rewrite the integrand:

      2x+1x+2=2xx+2+1x+2\frac{2 x + 1}{x + 2} = \frac{2 x}{x + 2} + \frac{1}{x + 2}

    2. Integrate term-by-term:

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

        2xx+2dx=2xx+2dx\int \frac{2 x}{x + 2}\, dx = 2 \int \frac{x}{x + 2}\, dx

        1. Rewrite the integrand:

          xx+2=12x+2\frac{x}{x + 2} = 1 - \frac{2}{x + 2}

        2. Integrate term-by-term:

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

            1dx=x\int 1\, dx = x

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

            (2x+2)dx=21x+2dx\int \left(- \frac{2}{x + 2}\right)\, dx = - 2 \int \frac{1}{x + 2}\, dx

            1. Let u=x+2u = x + 2.

              Then let du=dxdu = dx and substitute dudu:

              1udu\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+2)\log{\left(x + 2 \right)}

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

          The result is: x2log(x+2)x - 2 \log{\left(x + 2 \right)}

        So, the result is: 2x4log(x+2)2 x - 4 \log{\left(x + 2 \right)}

      1. Let u=x+2u = x + 2.

        Then let du=dxdu = dx and substitute dudu:

        1udu\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+2)\log{\left(x + 2 \right)}

      The result is: 2x+log(x+2)4log(x+2)2 x + \log{\left(x + 2 \right)} - 4 \log{\left(x + 2 \right)}

  2. Add the constant of integration:

    2x3log(2x+4)+4+constant2 x - 3 \log{\left(2 x + 4 \right)} + 4+ \mathrm{constant}


The answer is:

2x3log(2x+4)+4+constant2 x - 3 \log{\left(2 x + 4 \right)} + 4+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                         
 |                                          
 | 2*x + 1                                  
 | ------- dx = 4 + C - 3*log(4 + 2*x) + 2*x
 |  x + 2                                   
 |                                          
/                                           
2x+1x+2dx=C+2x3log(2x+4)+4\int \frac{2 x + 1}{x + 2}\, dx = C + 2 x - 3 \log{\left(2 x + 4 \right)} + 4
The graph
0.001.000.100.200.300.400.500.600.700.800.905-5
The answer [src]
2 - 3*log(3) + 3*log(2)
3log(3)+2+3log(2)- 3 \log{\left(3 \right)} + 2 + 3 \log{\left(2 \right)}
=
=
2 - 3*log(3) + 3*log(2)
3log(3)+2+3log(2)- 3 \log{\left(3 \right)} + 2 + 3 \log{\left(2 \right)}
2 - 3*log(3) + 3*log(2)
Numerical answer [src]
0.783604675675507
0.783604675675507
The graph
Integral of (2*x+1)/(x+2) dx

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