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1/(3x+1)
Solving integrals/ 1/(3x+1)

Integral of 1/(3x+1) dx

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Piecewise:

The solution

You have entered [src] 
  1           
  /           
 |            
 |     1      
 |  ------- dx
 |  3*x + 1   
 |            
/             
0
∫0113x+1 dx\int\limits_{0}^{1} \frac{1}{3 x + 1}\, dx 
Integral(1/(3*x + 1), (x, 0, 1))
Detail solution

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

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

    ∫13u du\int \frac{1}{3 u}\, du

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

      ∫1u du=∫1u du3\int \frac{1}{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}

  2. Now simplify:

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

  3. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                             
 |                              
 |    1             log(3*x + 1)
 | ------- dx = C + ------------
 | 3*x + 1               3      
 |                              
/
∫13x+1 dx=C+log⁡(3x+1)3\int \frac{1}{3 x + 1}\, dx = C + \frac{\log{\left(3 x + 1 \right)}}{3}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9002
Plot the graph f(x)
The answer [src] 
log(4)
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  3
log⁡(4)3\frac{\log{\left(4 \right)}}{3} 
=
= 
log(4)
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  3
log⁡(4)3\frac{\log{\left(4 \right)}}{3} 
log(4)/3
Numerical answer [src] 
0.462098120373297
0.462098120373297
The graph
Integral of 1/(3x+1) dx

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

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