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Solving integrals/ x*ln(1+x/1-x)

Integral of x*ln(1+x/1-x) dx

Limits of integration:

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v

The graph:

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Piecewise:

The solution

You have entered [src] 
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 |       /    x    \   
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∫01xlog⁡(−x+(x1+1)) dx\int\limits_{0}^{1} x \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}\, dx 
Integral(x*log(1 + x/1 - x), (x, 0, 1))
Detail solution

  1. Use integration by parts, noting that the integrand eventually repeats itself.

    1. For the integrand xlog⁡(−x+(x1+1))x \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}:

      Let u(x)=log⁡(−x+(x1+1))u{\left(x \right)} = \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)} and let dv⁡(x)=x\operatorname{dv}{\left(x \right)} = x.

      Then ∫xlog⁡(−x+(x1+1)) dx=x2log⁡(−x+(x1+1))2−∫0 dx\int x \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}\, dx = \frac{x^{2} \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}}{2} - \int 0\, dx.

    2. Notice that the integrand has repeated itself, so move it to one side:

      ∫xlog⁡(−x+(x1+1)) dx=x2log⁡(−x+(x1+1))2\int x \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}\, dx = \frac{x^{2} \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}}{2}

      Therefore,

      ∫xlog⁡(−x+(x1+1)) dx=x2log⁡(−x+(x1+1))2\int x \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}\, dx = \frac{x^{2} \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}}{2}

  2. Now simplify:

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  3. Add the constant of integration:

    0+constant0+ \mathrm{constant}

The answer is:

0+constant0+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                           2    /    x    \
 |                           x *log|1 + - - x|
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 | x*log|1 + - - x| dx = C + -----------------
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∫xlog⁡(−x+(x1+1)) dx=C+x2log⁡(−x+(x1+1))2\int x \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}\, dx = C + \frac{x^{2} \log{\left(- x + \left(\frac{x}{1} + 1\right) \right)}}{2}
Simplify
The graph
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Plot the graph f(x)
The answer [src] 
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Numerical answer [src] 
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    Use the examples entering the upper and lower limits of integration.

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