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(x+4)/(x+1)
Solving integrals/ (x+4)/(x+1)

Integral of (x+4)/(x+1) dx

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The solution

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 |  x + 4   
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∫01x+4x+1 dx\int\limits_{0}^{1} \frac{x + 4}{x + 1}\, dx 
Integral((x + 4)/(x + 1), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      x+4x+1=1+3x+1\frac{x + 4}{x + 1} = 1 + \frac{3}{x + 1}

    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+1 dx=3∫1x+1 dx\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:

          ∫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+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)x + 3 \log{\left(x + 1 \right)}

    Method #2

    1. Rewrite the integrand:

      x+4x+1=xx+1+4x+1\frac{x + 4}{x + 1} = \frac{x}{x + 1} + \frac{4}{x + 1}

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        xx+1=1−1x+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:

          ∫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:

          ∫(−1x+1) dx=−∫1x+1 dx\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:

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

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

        The result is: x−log⁡(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:

        ∫4x+1 dx=4∫1x+1 dx\int \frac{4}{x + 1}\, dx = 4 \int \frac{1}{x + 1}\, dx

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

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

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

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

  2. Add the constant of integration:

    x+3log⁡(x+1)+constantx + 3 \log{\left(x + 1 \right)}+ \mathrm{constant}

The answer is:

x+3log⁡(x+1)+constantx + 3 \log{\left(x + 1 \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                               
 |                                
 | x + 4                          
 | ----- dx = C + x + 3*log(1 + x)
 | x + 1                          
 |                                
/
3 log⁡(x+1)+x3\,\log \left(x+1\right)+x
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9005
Plot the graph f(x)
The answer [src] 
1 + 3*log(2)
3 log⁡2+13\,\log 2+1 
=
= 
1 + 3*log(2)
1+3log⁡(2)1 + 3 \log{\left(2 \right)}
Simplify
Numerical answer [src] 
3.07944154167984
3.07944154167984
The graph
Integral of (x+4)/(x+1) dx

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

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