Mister Exam

Integral of ln(3x+1) dx

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The solution

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01log(3x+1)dx\int\limits_{0}^{1} \log{\left(3 x + 1 \right)}\, dx
Integral(log(3*x + 1), (x, 0, 1))
Detail solution
  1. There are multiple ways to do this integral.

    Method #1

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

      Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

      log(u)9du\int \frac{\log{\left(u \right)}}{9}\, du

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

        log(u)3du=log(u)du3\int \frac{\log{\left(u \right)}}{3}\, du = \frac{\int \log{\left(u \right)}\, du}{3}

        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

        So, the result is: ulog(u)3u3\frac{u \log{\left(u \right)}}{3} - \frac{u}{3}

      Now substitute uu back in:

      x+(3x+1)log(3x+1)313- x + \frac{\left(3 x + 1\right) \log{\left(3 x + 1 \right)}}{3} - \frac{1}{3}

    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(3x+1)u{\left(x \right)} = \log{\left(3 x + 1 \right)} and let dv(x)=1\operatorname{dv}{\left(x \right)} = 1.

      Then du(x)=33x+1\operatorname{du}{\left(x \right)} = \frac{3}{3 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. The integral of a constant times a function is the constant times the integral of the function:

      3x3x+1dx=3x3x+1dx\int \frac{3 x}{3 x + 1}\, dx = 3 \int \frac{x}{3 x + 1}\, dx

      1. Rewrite the integrand:

        x3x+1=1313(3x+1)\frac{x}{3 x + 1} = \frac{1}{3} - \frac{1}{3 \cdot \left(3 x + 1\right)}

      2. Integrate term-by-term:

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

          13dx=x3\int \frac{1}{3}\, dx = \frac{x}{3}

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

          (13(3x+1))dx=13x+1dx3\int \left(- \frac{1}{3 \cdot \left(3 x + 1\right)}\right)\, dx = - \frac{\int \frac{1}{3 x + 1}\, dx}{3}

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

            Then let du=3dxdu = 3 dx and substitute du3\frac{du}{3}:

            19udu\int \frac{1}{9 u}\, du

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

              13udu=1udu3\int \frac{1}{3 u}\, du = \frac{\int \frac{1}{u}\, du}{3}

              1. The integral of 1u\frac{1}{u} is log(u)\log{\left(u \right)}.

              So, the result is: log(u)3\frac{\log{\left(u \right)}}{3}

            Now substitute uu back in:

            log(3x+1)3\frac{\log{\left(3 x + 1 \right)}}{3}

          So, the result is: log(3x+1)9- \frac{\log{\left(3 x + 1 \right)}}{9}

        The result is: x3log(3x+1)9\frac{x}{3} - \frac{\log{\left(3 x + 1 \right)}}{9}

      So, the result is: xlog(3x+1)3x - \frac{\log{\left(3 x + 1 \right)}}{3}

  2. Now simplify:

    x+(3x+1)log(3x+1)313- x + \frac{\left(3 x + 1\right) \log{\left(3 x + 1 \right)}}{3} - \frac{1}{3}

  3. Add the constant of integration:

    x+(3x+1)log(3x+1)313+constant- x + \frac{\left(3 x + 1\right) \log{\left(3 x + 1 \right)}}{3} - \frac{1}{3}+ \mathrm{constant}


The answer is:

x+(3x+1)log(3x+1)313+constant- x + \frac{\left(3 x + 1\right) \log{\left(3 x + 1 \right)}}{3} - \frac{1}{3}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                                      
 |                     1           (3*x + 1)*log(3*x + 1)
 | log(3*x + 1) dx = - - + C - x + ----------------------
 |                     3                     3           
/                                                        
(3x+1)log(3x+1)3x13{{\left(3\,x+1\right)\,\log \left(3\,x+1\right)-3\,x-1}\over{3}}
The graph
0.001.000.100.200.300.400.500.600.700.800.9002
The answer [src]
     4*log(4)
-1 + --------
        3    
4log433{{4\,\log 4-3}\over{3}}
=
=
     4*log(4)
-1 + --------
        3    
1+4log(4)3-1 + \frac{4 \log{\left(4 \right)}}{3}
Numerical answer [src]
0.848392481493187
0.848392481493187
The graph
Integral of ln(3x+1) dx

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