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(x+2)/(x+3)

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

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The solution

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  1         
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 |  x + 2   
 |  ----- dx
 |  x + 3   
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01x+2x+3dx\int\limits_{0}^{1} \frac{x + 2}{x + 3}\, dx
Integral((x + 2)/(x + 3), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      x+2x+3=11x+3\frac{x + 2}{x + 3} = 1 - \frac{1}{x + 3}

    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:

        (1x+3)dx=1x+3dx\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:

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

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

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

    Method #2

    1. Rewrite the integrand:

      x+2x+3=xx+3+2x+3\frac{x + 2}{x + 3} = \frac{x}{x + 3} + \frac{2}{x + 3}

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        xx+3=13x+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:

          1dx=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=31x+3dx\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:

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

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

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

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

        2x+3dx=21x+3dx\int \frac{2}{x + 3}\, dx = 2 \int \frac{1}{x + 3}\, dx

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

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

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

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

  2. Add the constant of integration:

    xlog(x+3)+constantx - \log{\left(x + 3 \right)}+ \mathrm{constant}


The answer is:

xlog(x+3)+constantx - \log{\left(x + 3 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                             
 |                              
 | x + 2                        
 | ----- dx = C + x - log(3 + x)
 | x + 3                        
 |                              
/                               
x+2x+3dx=C+xlog(x+3)\int \frac{x + 2}{x + 3}\, dx = C + x - \log{\left(x + 3 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.902-2
The answer [src]
1 - log(4) + log(3)
log(4)+1+log(3)- \log{\left(4 \right)} + 1 + \log{\left(3 \right)}
=
=
1 - log(4) + log(3)
log(4)+1+log(3)- \log{\left(4 \right)} + 1 + \log{\left(3 \right)}
1 - log(4) + log(3)
Numerical answer [src]
0.712317927548219
0.712317927548219
The graph
Integral of (x+2)/(x+3) dx

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