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log(x+1)/(x+1)
Solving integrals/ log(x+1)/(x+1)

You entered:

log(x+1)/(x+1)

What you mean?

log(x + 1)/(x + 1)
 
log(x + 1)/x + 1
 
log(x + 1)/(x + 1)
 
log(x + 1)/x + 1
 
log(x + 1)/(x + 1)
 
log(x + 1)/x + 1
 
log(x + 1)/x + 1

Integral of log(x+1)/(x+1) dx

Limits of integration:

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v

The graph:

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Piecewise:

The solution

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

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=log⁡(x+1)u = \log{\left(x + 1 \right)}.

      Then let du=dxx+1du = \frac{dx}{x + 1} and substitute dudu:

      ∫u du\int u\, du

      1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫u du=u22\int u\, du = \frac{u^{2}}{2}

      Now substitute uu back in:

      log⁡(x+1)22\frac{\log{\left(x + 1 \right)}^{2}}{2}

    Method #2

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

      Then let du=dxdu = dx and substitute dudu:

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

      1. Let u=1uu = \frac{1}{u}.

        Then let du=−duu2du = - \frac{du}{u^{2}} and substitute −du- du:

        ∫log⁡(1u)u du\int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du

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

          ∫(−log⁡(1u)u) du=−∫log⁡(1u)u du\int \left(- \frac{\log{\left(\frac{1}{u} \right)}}{u}\right)\, du = - \int \frac{\log{\left(\frac{1}{u} \right)}}{u}\, du

          1. Let u=log⁡(1u)u = \log{\left(\frac{1}{u} \right)}.

            Then let du=−duudu = - \frac{du}{u} and substitute −du- du:

            ∫u du\int u\, du

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

              ∫(−u) du=−∫u du\int \left(- u\right)\, du = - \int u\, du

              1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

                ∫u du=u22\int u\, du = \frac{u^{2}}{2}

              So, the result is: −u22- \frac{u^{2}}{2}

            Now substitute uu back in:

            −log⁡(1u)22- \frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

          So, the result is: log⁡(1u)22\frac{\log{\left(\frac{1}{u} \right)}^{2}}{2}

        Now substitute uu back in:

        log⁡(u)22\frac{\log{\left(u \right)}^{2}}{2}

      Now substitute uu back in:

      log⁡(x+1)22\frac{\log{\left(x + 1 \right)}^{2}}{2}

  2. Now simplify:

    log⁡(x+1)22\frac{\log{\left(x + 1 \right)}^{2}}{2}

  3. Add the constant of integration:

    log⁡(x+1)22+constant\frac{\log{\left(x + 1 \right)}^{2}}{2}+ \mathrm{constant}

The answer is:

log⁡(x+1)22+constant\frac{\log{\left(x + 1 \right)}^{2}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                               
 |                        2       
 | log(x + 1)          log (x + 1)
 | ---------- dx = C + -----------
 |   x + 1                  2     
 |                                
/
(log⁡(x+1))22{{\left(\log \left(x+1\right)\right)^2}\over{2}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.5
Plot the graph f(x)
The answer [src] 
   2   
log (2)
-------
   2
(log⁡2)22{{\left(\log 2\right)^2}\over{2}} 
=
= 
   2   
log (2)
-------
   2
log⁡(2)22\frac{\log{\left(2 \right)}^{2}}{2}
Simplify
Numerical answer [src] 
0.240226506959101
0.240226506959101
The graph
Integral of log(x+1)/(x+1) dx

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

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