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Solving integrals/ log(2*x+3)/(2*x+3)

Integral of log(2*x+3)/(2*x+3) dx

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The solution

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  1                
  /                
 |                 
 |  log(2*x + 3)   
 |  ------------ dx
 |    2*x + 3      
 |                 
/                  
0
∫01log⁡(2x+3)2x+3 dx\int\limits_{0}^{1} \frac{\log{\left(2 x + 3 \right)}}{2 x + 3}\, dx 
Integral(log(2*x + 3)/(2*x + 3), (x, 0, 1))
Detail solution

  1. Let u=2x+3u = 2 x + 3.

    Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

    ∫log⁡(u)2u du\int \frac{\log{\left(u \right)}}{2 u}\, du

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

      ∫log⁡(u)u du=∫log⁡(u)u du2\int \frac{\log{\left(u \right)}}{u}\, du = \frac{\int \frac{\log{\left(u \right)}}{u}\, du}{2}

      1. There are multiple ways to do this integral.

        Method #1

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

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

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

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

                ∫u du=−∫u du\int 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}

                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}

        Method #2

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

          Then let du=duudu = \frac{du}{u} 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⁡(u)22\frac{\log{\left(u \right)}^{2}}{2}

      So, the result is: log⁡(u)24\frac{\log{\left(u \right)}^{2}}{4}

    Now substitute uu back in:

    log⁡(2x+3)24\frac{\log{\left(2 x + 3 \right)}^{2}}{4}

  2. Now simplify:

    log⁡(2x+3)24\frac{\log{\left(2 x + 3 \right)}^{2}}{4}

  3. Add the constant of integration:

    log⁡(2x+3)24+constant\frac{\log{\left(2 x + 3 \right)}^{2}}{4}+ \mathrm{constant}

The answer is:

log⁡(2x+3)24+constant\frac{\log{\left(2 x + 3 \right)}^{2}}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
 |                          2         
 | log(2*x + 3)          log (2*x + 3)
 | ------------ dx = C + -------------
 |   2*x + 3                   4      
 |                                    
/
∫log⁡(2x+3)2x+3 dx=C+log⁡(2x+3)24\int \frac{\log{\left(2 x + 3 \right)}}{2 x + 3}\, dx = C + \frac{\log{\left(2 x + 3 \right)}^{2}}{4}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.01.0
Plot the graph f(x)
The answer [src] 
     2         2   
  log (3)   log (5)
- ------- + -------
     4         4
−log⁡(3)24+log⁡(5)24- \frac{\log{\left(3 \right)}^{2}}{4} + \frac{\log{\left(5 \right)}^{2}}{4} 
=
= 
     2         2   
  log (3)   log (5)
- ------- + -------
     4         4
−log⁡(3)24+log⁡(5)24- \frac{\log{\left(3 \right)}^{2}}{4} + \frac{\log{\left(5 \right)}^{2}}{4} 
-log(3)^2/4 + log(5)^2/4
Numerical answer [src] 
0.345835358291913
0.345835358291913

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

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