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(x+3)/(x+1)

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

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The solution

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

    Method #1

    1. Rewrite the integrand:

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

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

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

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

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

      The result is: x+2log(x+1)x + 2 \log{\left(x + 1 \right)}

    Method #2

    1. Rewrite the integrand:

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

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        xx+1=11x+1\frac{x}{x + 1} = 1 - \frac{1}{x + 1}

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

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

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

          So, the result is: log(x+1)- \log{\left(x + 1 \right)}

        The result is: xlog(x+1)x - \log{\left(x + 1 \right)}

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

        3x+1dx=31x+1dx\int \frac{3}{x + 1}\, dx = 3 \int \frac{1}{x + 1}\, dx

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

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

        So, the result is: 3log(x+1)3 \log{\left(x + 1 \right)}

      The result is: x+3log(x+1)log(x+1)x + 3 \log{\left(x + 1 \right)} - \log{\left(x + 1 \right)}

  2. Add the constant of integration:

    x+2log(x+1)+constantx + 2 \log{\left(x + 1 \right)}+ \mathrm{constant}


The answer is:

x+2log(x+1)+constantx + 2 \log{\left(x + 1 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                               
 |                                
 | x + 3                          
 | ----- dx = C + x + 2*log(1 + x)
 | x + 1                          
 |                                
/                                 
x+3x+1dx=C+x+2log(x+1)\int \frac{x + 3}{x + 1}\, dx = C + x + 2 \log{\left(x + 1 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.9005
The answer [src]
1 + 2*log(2)
1+2log(2)1 + 2 \log{\left(2 \right)}
=
=
1 + 2*log(2)
1+2log(2)1 + 2 \log{\left(2 \right)}
1 + 2*log(2)
Numerical answer [src]
2.38629436111989
2.38629436111989
The graph
Integral of (x+3)/(x+1) dx

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