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ln(x+1)

Integral of ln(x+1) dx

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The solution

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01log(x+1)dx\int\limits_{0}^{1} \log{\left(x + 1 \right)}\, dx
Integral(log(x + 1), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=dxdu = dx and substitute dudu:

      log(u)du\int \log{\left(u \right)}\, du

      1. Use integration by parts:

        udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

        Let u(u)=log(u)u{\left(u \right)} = \log{\left(u \right)} and let dv(u)=1\operatorname{dv}{\left(u \right)} = 1.

        Then du(u)=1u\operatorname{du}{\left(u \right)} = \frac{1}{u}.

        To find v(u)v{\left(u \right)}:

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

          1du=u\int 1\, du = u

        Now evaluate the sub-integral.

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

        1du=u\int 1\, du = u

      Now substitute uu back in:

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

    Method #2

    1. Use integration by parts:

      udv=uvvdu\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

      Let u(x)=log(x+1)u{\left(x \right)} = \log{\left(x + 1 \right)} and let dv(x)=1\operatorname{dv}{\left(x \right)} = 1.

      Then du(x)=1x+1\operatorname{du}{\left(x \right)} = \frac{1}{x + 1}.

      To find v(x)v{\left(x \right)}:

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

        1dx=x\int 1\, dx = x

      Now evaluate the sub-integral.

    2. Rewrite the integrand:

      xx+1=11x+1\frac{x}{x + 1} = 1 - \frac{1}{x + 1}

    3. 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)}

  2. Now simplify:

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

  3. Add the constant of integration:

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


The answer is:

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

The answer (Indefinite) [src]
  /                                               
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 | log(x + 1) dx = -1 + C - x + (x + 1)*log(x + 1)
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log(x+1)dx=Cx+(x+1)log(x+1)1\int \log{\left(x + 1 \right)}\, dx = C - x + \left(x + 1\right) \log{\left(x + 1 \right)} - 1
The graph
0.001.000.100.200.300.400.500.600.700.800.900.01.0
The answer [src]
-1 + 2*log(2)
1+2log(2)-1 + 2 \log{\left(2 \right)}
=
=
-1 + 2*log(2)
1+2log(2)-1 + 2 \log{\left(2 \right)}
Numerical answer [src]
0.386294361119891
0.386294361119891
The graph
Integral of ln(x+1) dx

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