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x/(1+x)

Integral of x/(1+x) dx

Limits of integration:

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The solution

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01xx+1dx\int\limits_{0}^{1} \frac{x}{x + 1}\, dx
Integral(x/(1 + x), (x, 0, 1))
Detail solution
  1. Rewrite the integrand:

    xx+1=11x+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:

      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+1)dx=1x+1dx\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:

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

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

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

  3. Add the constant of integration:

    xlog(x+1)+constantx - \log{\left(x + 1 \right)}+ \mathrm{constant}


The answer is:

xlog(x+1)+constantx - \log{\left(x + 1 \right)}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                             
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 | 1 + x                        
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xx+1dx=C+xlog(x+1)\int \frac{x}{x + 1}\, dx = C + x - \log{\left(x + 1 \right)}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.01.0
The answer [src]
1 - log(2)
1log(2)1 - \log{\left(2 \right)}
=
=
1 - log(2)
1log(2)1 - \log{\left(2 \right)}
1 - log(2)
Numerical answer [src]
0.306852819440055
0.306852819440055
The graph
Integral of x/(1+x) dx

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