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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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  1         
  /         
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 |  x + 3   
 |  ----- dx
 |  x - 1   
 |          
/           
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01x+3x1dx\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+3x1=1+4x1\frac{x + 3}{x - 1} = 1 + \frac{4}{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:

        4x1dx=41x1dx\int \frac{4}{x - 1}\, dx = 4 \int \frac{1}{x - 1}\, dx

        1. Let u=x1u = 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(x1)\log{\left(x - 1 \right)}

        So, the result is: 4log(x1)4 \log{\left(x - 1 \right)}

      The result is: x+4log(x1)x + 4 \log{\left(x - 1 \right)}

    Method #2

    1. Rewrite the integrand:

      x+3x1=xx1+3x1\frac{x + 3}{x - 1} = \frac{x}{x - 1} + \frac{3}{x - 1}

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        xx1=1+1x1\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. Let u=x1u = 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(x1)\log{\left(x - 1 \right)}

        The result is: x+log(x1)x + \log{\left(x - 1 \right)}

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

        3x1dx=31x1dx\int \frac{3}{x - 1}\, dx = 3 \int \frac{1}{x - 1}\, dx

        1. Let u=x1u = 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(x1)\log{\left(x - 1 \right)}

        So, the result is: 3log(x1)3 \log{\left(x - 1 \right)}

      The result is: x+log(x1)+3log(x1)x + \log{\left(x - 1 \right)} + 3 \log{\left(x - 1 \right)}

  2. Add the constant of integration:

    x+4log(x1)+constantx + 4 \log{\left(x - 1 \right)}+ \mathrm{constant}


The answer is:

x+4log(x1)+constantx + 4 \log{\left(x - 1 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                
 |                                 
 | x + 3                           
 | ----- dx = C + x + 4*log(-1 + x)
 | x - 1                           
 |                                 
/                                  
x+3x1dx=C+x+4log(x1)\int \frac{x + 3}{x - 1}\, dx = C + x + 4 \log{\left(x - 1 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.90-5000050000
The answer [src]
-oo - 4*pi*I
4iπ-\infty - 4 i \pi
=
=
-oo - 4*pi*I
4iπ-\infty - 4 i \pi
-oo - 4*pi*i
Numerical answer [src]
-175.363827144878
-175.363827144878
The graph
Integral of (x+3)/(x-1) dx

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