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Solving integrals/ 1/(sqrt(3x+1)+2)

Integral of 1/(sqrt(3x+1)+2) dx

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The solution

You have entered [src] 
  1                   
  /                   
 |                    
 |         1          
 |  --------------- dx
 |    _________       
 |  \/ 3*x + 1  + 2   
 |                    
/                     
0
∫0113x+1+2 dx\int\limits_{0}^{1} \frac{1}{\sqrt{3 x + 1} + 2}\, dx 
Integral(1/(sqrt(3*x + 1) + 2), (x, 0, 1))
Detail solution

  1. Let u=3x+1u = \sqrt{3 x + 1}.

    Then let du=3dx23x+1du = \frac{3 dx}{2 \sqrt{3 x + 1}} and substitute 2du2 du:

    ∫2u3u+6 du\int \frac{2 u}{3 u + 6}\, du

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

      ∫u3u+6 du=2∫u3u+6 du\int \frac{u}{3 u + 6}\, du = 2 \int \frac{u}{3 u + 6}\, du

      1. Rewrite the integrand:

        u3u+6=13−23(u+2)\frac{u}{3 u + 6} = \frac{1}{3} - \frac{2}{3 \left(u + 2\right)}

      2. Integrate term-by-term:

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

          ∫13 du=u3\int \frac{1}{3}\, du = \frac{u}{3}

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

          ∫(−23(u+2)) du=−2∫1u+2 du3\int \left(- \frac{2}{3 \left(u + 2\right)}\right)\, du = - \frac{2 \int \frac{1}{u + 2}\, du}{3}

          1. Let u=u+2u = u + 2.

            Then let du=dudu = du and substitute dudu:

            ∫1u du\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⁡(u+2)\log{\left(u + 2 \right)}

          So, the result is: −2log⁡(u+2)3- \frac{2 \log{\left(u + 2 \right)}}{3}

        The result is: u3−2log⁡(u+2)3\frac{u}{3} - \frac{2 \log{\left(u + 2 \right)}}{3}

      So, the result is: 2u3−4log⁡(u+2)3\frac{2 u}{3} - \frac{4 \log{\left(u + 2 \right)}}{3}

    Now substitute uu back in:

    23x+13−4log⁡(3x+1+2)3\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}

  2. Now simplify:

    23x+13−4log⁡(3x+1+2)3\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}

  3. Add the constant of integration:

    23x+13−4log⁡(3x+1+2)3+constant\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}+ \mathrm{constant}

The answer is:

23x+13−4log⁡(3x+1+2)3+constant\frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                               
 |                               /      _________\       _________
 |        1                 4*log\2 + \/ 3*x + 1 /   2*\/ 3*x + 1 
 | --------------- dx = C - ---------------------- + -------------
 |   _________                        3                    3      
 | \/ 3*x + 1  + 2                                                
 |                                                                
/
∫13x+1+2 dx=C+23x+13−4log⁡(3x+1+2)3\int \frac{1}{\sqrt{3 x + 1} + 2}\, dx = C + \frac{2 \sqrt{3 x + 1}}{3} - \frac{4 \log{\left(\sqrt{3 x + 1} + 2 \right)}}{3}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.901-1
Plot the graph f(x)
The answer [src] 
2   4*log(4)   4*log(3)
- - -------- + --------
3      3          3
−4log⁡(4)3+23+4log⁡(3)3- \frac{4 \log{\left(4 \right)}}{3} + \frac{2}{3} + \frac{4 \log{\left(3 \right)}}{3} 
=
= 
2   4*log(4)   4*log(3)
- - -------- + --------
3      3          3
−4log⁡(4)3+23+4log⁡(3)3- \frac{4 \log{\left(4 \right)}}{3} + \frac{2}{3} + \frac{4 \log{\left(3 \right)}}{3} 
2/3 - 4*log(4)/3 + 4*log(3)/3
Numerical answer [src] 
0.283090570064292
0.283090570064292

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

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