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(x^2+1)/(x+2)

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

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The solution

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01x2+1x+2dx\int\limits_{0}^{1} \frac{x^{2} + 1}{x + 2}\, dx
Integral((x^2 + 1)/(x + 2), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      x2+1x+2=x2+5x+2\frac{x^{2} + 1}{x + 2} = x - 2 + \frac{5}{x + 2}

    2. Integrate term-by-term:

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n1n \neq -1:

        xdx=x22\int x\, dx = \frac{x^{2}}{2}

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

        (2)dx=2x\int \left(-2\right)\, dx = - 2 x

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

        5x+2dx=51x+2dx\int \frac{5}{x + 2}\, dx = 5 \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: 5log(x+2)5 \log{\left(x + 2 \right)}

      The result is: x222x+5log(x+2)\frac{x^{2}}{2} - 2 x + 5 \log{\left(x + 2 \right)}

    Method #2

    1. Rewrite the integrand:

      x2+1x+2=x2x+2+1x+2\frac{x^{2} + 1}{x + 2} = \frac{x^{2}}{x + 2} + \frac{1}{x + 2}

    2. Integrate term-by-term:

      1. Rewrite the integrand:

        x2x+2=x2+4x+2\frac{x^{2}}{x + 2} = x - 2 + \frac{4}{x + 2}

      2. Integrate term-by-term:

        1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n1n \neq -1:

          xdx=x22\int x\, dx = \frac{x^{2}}{2}

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

          (2)dx=2x\int \left(-2\right)\, dx = - 2 x

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

          4x+2dx=41x+2dx\int \frac{4}{x + 2}\, dx = 4 \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: 4log(x+2)4 \log{\left(x + 2 \right)}

        The result is: x222x+4log(x+2)\frac{x^{2}}{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: x222x+log(x+2)+4log(x+2)\frac{x^{2}}{2} - 2 x + \log{\left(x + 2 \right)} + 4 \log{\left(x + 2 \right)}

  2. Add the constant of integration:

    x222x+5log(x+2)+constant\frac{x^{2}}{2} - 2 x + 5 \log{\left(x + 2 \right)}+ \mathrm{constant}


The answer is:

x222x+5log(x+2)+constant\frac{x^{2}}{2} - 2 x + 5 \log{\left(x + 2 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                       
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 |  2               2                     
 | x  + 1          x                      
 | ------ dx = C + -- - 2*x + 5*log(2 + x)
 | x + 2           2                      
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/                                         
x2+1x+2dx=C+x222x+5log(x+2)\int \frac{x^{2} + 1}{x + 2}\, dx = C + \frac{x^{2}}{2} - 2 x + 5 \log{\left(x + 2 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.9005
The answer [src]
-3/2 - 5*log(2) + 5*log(3)
5log(2)32+5log(3)- 5 \log{\left(2 \right)} - \frac{3}{2} + 5 \log{\left(3 \right)}
=
=
-3/2 - 5*log(2) + 5*log(3)
5log(2)32+5log(3)- 5 \log{\left(2 \right)} - \frac{3}{2} + 5 \log{\left(3 \right)}
-3/2 - 5*log(2) + 5*log(3)
Numerical answer [src]
0.527325540540822
0.527325540540822
The graph
Integral of (x^2+1)/(x+2) dx

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