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x^2/(4-x^2)

Integral of x^2/(4-x^2) dx

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The solution

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01x24x2dx\int\limits_{0}^{1} \frac{x^{2}}{4 - x^{2}}\, dx
Integral(x^2/(4 - x^2), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

    1. Rewrite the integrand:

      x24x2=1+1x+21x2\frac{x^{2}}{4 - x^{2}} = -1 + \frac{1}{x + 2} - \frac{1}{x - 2}

    2. Integrate term-by-term:

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

        (1)dx=x\int \left(-1\right)\, dx = - x

      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)}

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

        (1x2)dx=1x2dx\int \left(- \frac{1}{x - 2}\right)\, dx = - \int \frac{1}{x - 2}\, dx

        1. Let u=x2u = 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(x2)\log{\left(x - 2 \right)}

        So, the result is: log(x2)- \log{\left(x - 2 \right)}

      The result is: xlog(x2)+log(x+2)- x - \log{\left(x - 2 \right)} + \log{\left(x + 2 \right)}

    Method #2

    1. Rewrite the integrand:

      x24x2=x2x24\frac{x^{2}}{4 - x^{2}} = - \frac{x^{2}}{x^{2} - 4}

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

      (x2x24)dx=x2x24dx\int \left(- \frac{x^{2}}{x^{2} - 4}\right)\, dx = - \int \frac{x^{2}}{x^{2} - 4}\, dx

      1. Rewrite the integrand:

        x2x24=11x+2+1x2\frac{x^{2}}{x^{2} - 4} = 1 - \frac{1}{x + 2} + \frac{1}{x - 2}

      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+2)dx=1x+2dx\int \left(- \frac{1}{x + 2}\right)\, dx = - \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: log(x+2)- \log{\left(x + 2 \right)}

        1. Let u=x2u = 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(x2)\log{\left(x - 2 \right)}

        The result is: x+log(x2)log(x+2)x + \log{\left(x - 2 \right)} - \log{\left(x + 2 \right)}

      So, the result is: xlog(x2)+log(x+2)- x - \log{\left(x - 2 \right)} + \log{\left(x + 2 \right)}

  2. Add the constant of integration:

    xlog(x2)+log(x+2)+constant- x - \log{\left(x - 2 \right)} + \log{\left(x + 2 \right)}+ \mathrm{constant}


The answer is:

xlog(x2)+log(x+2)+constant- x - \log{\left(x - 2 \right)} + \log{\left(x + 2 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                            
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/                                              
x24x2dx=Cxlog(x2)+log(x+2)\int \frac{x^{2}}{4 - x^{2}}\, dx = C - x - \log{\left(x - 2 \right)} + \log{\left(x + 2 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.5
The answer [src]
-1 + log(3)
1+log(3)-1 + \log{\left(3 \right)}
=
=
-1 + log(3)
1+log(3)-1 + \log{\left(3 \right)}
-1 + log(3)
Numerical answer [src]
0.0986122886681097
0.0986122886681097
The graph
Integral of x^2/(4-x^2) dx

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