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x^3/(x^2-4)
Solving integrals/ x^3/(x^2-4)

Integral of x^3/(x^2-4) dx

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The solution

You have entered [src] 
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∫01x3x2−4 dx\int\limits_{0}^{1} \frac{x^{3}}{x^{2} - 4}\, dx 
Integral(x^3/(x^2 - 4), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=x2u = x^{2}.

      Then let du=2xdxdu = 2 x dx and substitute dudu:

      ∫u2u−8 du\int \frac{u}{2 u - 8}\, du

      1. Rewrite the integrand:

        u2u−8=12+2u−4\frac{u}{2 u - 8} = \frac{1}{2} + \frac{2}{u - 4}

      2. Integrate term-by-term:

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

          ∫12 du=u2\int \frac{1}{2}\, du = \frac{u}{2}

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

          ∫2u−4 du=2∫1u−4 du\int \frac{2}{u - 4}\, du = 2 \int \frac{1}{u - 4}\, du

          1. Let u=u−4u = u - 4.

            Then let du=dudu = du 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⁡(u−4)\log{\left(u - 4 \right)}

          So, the result is: 2log⁡(u−4)2 \log{\left(u - 4 \right)}

        The result is: u2+2log⁡(u−4)\frac{u}{2} + 2 \log{\left(u - 4 \right)}

      Now substitute uu back in:

      x22+2log⁡(x2−4)\frac{x^{2}}{2} + 2 \log{\left(x^{2} - 4 \right)}

    Method #2

    1. Rewrite the integrand:

      x3x2−4=x+2x+2+2x−2\frac{x^{3}}{x^{2} - 4} = x + \frac{2}{x + 2} + \frac{2}{x - 2}

    2. Integrate term-by-term:

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

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

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

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

        1. Let u=x+2u = x + 2.

          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+2)\log{\left(x + 2 \right)}

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

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

        ∫2x−2 dx=2∫1x−2 dx\int \frac{2}{x - 2}\, dx = 2 \int \frac{1}{x - 2}\, dx

        1. Let u=x−2u = x - 2.

          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−2)\log{\left(x - 2 \right)}

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

      The result is: x22+2log⁡(x−2)+2log⁡(x+2)\frac{x^{2}}{2} + 2 \log{\left(x - 2 \right)} + 2 \log{\left(x + 2 \right)}

  2. Add the constant of integration:

    x22+2log⁡(x2−4)+constant\frac{x^{2}}{2} + 2 \log{\left(x^{2} - 4 \right)}+ \mathrm{constant}

The answer is:

x22+2log⁡(x2−4)+constant\frac{x^{2}}{2} + 2 \log{\left(x^{2} - 4 \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
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 |    3             2                 
 |   x             x         /      2\
 | ------ dx = C + -- + 2*log\-4 + x /
 |  2              2                  
 | x  - 4                             
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/
∫x3x2−4 dx=C+x22+2log⁡(x2−4)\int \frac{x^{3}}{x^{2} - 4}\, dx = C + \frac{x^{2}}{2} + 2 \log{\left(x^{2} - 4 \right)}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.5-0.5
Plot the graph f(x)
The answer [src] 
1/2 - 2*log(4) + 2*log(3)
−2log⁡(4)+12+2log⁡(3)- 2 \log{\left(4 \right)} + \frac{1}{2} + 2 \log{\left(3 \right)} 
=
= 
1/2 - 2*log(4) + 2*log(3)
−2log⁡(4)+12+2log⁡(3)- 2 \log{\left(4 \right)} + \frac{1}{2} + 2 \log{\left(3 \right)} 
1/2 - 2*log(4) + 2*log(3)
Numerical answer [src] 
-0.0753641449035619
-0.0753641449035619
The graph
Integral of x^3/(x^2-4) dx

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

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