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x/(sqrt(2x+1)+1)

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

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01x2x+1+1dx\int\limits_{0}^{1} \frac{x}{\sqrt{2 x + 1} + 1}\, dx
Integral(x/(sqrt(2*x + 1) + 1), (x, 0, 1))
Detail solution
  1. Let u=2x+1u = \sqrt{2 x + 1}.

    Then let du=dx2x+1du = \frac{dx}{\sqrt{2 x + 1}} and substitute dudu:

    u(u2212)u+1du\int \frac{u \left(\frac{u^{2}}{2} - \frac{1}{2}\right)}{u + 1}\, du

    1. There are multiple ways to do this integral.

      Method #1

      1. Rewrite the integrand:

        u(u2212)u+1=u22u2\frac{u \left(\frac{u^{2}}{2} - \frac{1}{2}\right)}{u + 1} = \frac{u^{2}}{2} - \frac{u}{2}

      2. Integrate term-by-term:

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

          u22du=u2du2\int \frac{u^{2}}{2}\, du = \frac{\int u^{2}\, du}{2}

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

            u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: u36\frac{u^{3}}{6}

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

          (u2)du=udu2\int \left(- \frac{u}{2}\right)\, du = - \frac{\int u\, du}{2}

          1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

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

          So, the result is: u24- \frac{u^{2}}{4}

        The result is: u36u24\frac{u^{3}}{6} - \frac{u^{2}}{4}

      Method #2

      1. Rewrite the integrand:

        u(u2212)u+1=u32(u+1)u2(u+1)\frac{u \left(\frac{u^{2}}{2} - \frac{1}{2}\right)}{u + 1} = \frac{u^{3}}{2 \left(u + 1\right)} - \frac{u}{2 \left(u + 1\right)}

      2. Integrate term-by-term:

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

          u32(u+1)du=u3u+1du2\int \frac{u^{3}}{2 \left(u + 1\right)}\, du = \frac{\int \frac{u^{3}}{u + 1}\, du}{2}

          1. Rewrite the integrand:

            u3u+1=u2u+11u+1\frac{u^{3}}{u + 1} = u^{2} - u + 1 - \frac{1}{u + 1}

          2. Integrate term-by-term:

            1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n1n \neq -1:

              u2du=u33\int u^{2}\, du = \frac{u^{3}}{3}

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

              (u)du=udu\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 n1n \neq -1:

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

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

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

              1du=u\int 1\, du = u

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

              (1u+1)du=1u+1du\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du

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

                Then let du=dudu = du and substitute dudu:

                1udu\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+1)\log{\left(u + 1 \right)}

              So, the result is: log(u+1)- \log{\left(u + 1 \right)}

            The result is: u33u22+ulog(u+1)\frac{u^{3}}{3} - \frac{u^{2}}{2} + u - \log{\left(u + 1 \right)}

          So, the result is: u36u24+u2log(u+1)2\frac{u^{3}}{6} - \frac{u^{2}}{4} + \frac{u}{2} - \frac{\log{\left(u + 1 \right)}}{2}

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

          (u2(u+1))du=uu+1du2\int \left(- \frac{u}{2 \left(u + 1\right)}\right)\, du = - \frac{\int \frac{u}{u + 1}\, du}{2}

          1. Rewrite the integrand:

            uu+1=11u+1\frac{u}{u + 1} = 1 - \frac{1}{u + 1}

          2. Integrate term-by-term:

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

              1du=u\int 1\, du = u

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

              (1u+1)du=1u+1du\int \left(- \frac{1}{u + 1}\right)\, du = - \int \frac{1}{u + 1}\, du

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

                Then let du=dudu = du and substitute dudu:

                1udu\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+1)\log{\left(u + 1 \right)}

              So, the result is: log(u+1)- \log{\left(u + 1 \right)}

            The result is: ulog(u+1)u - \log{\left(u + 1 \right)}

          So, the result is: u2+log(u+1)2- \frac{u}{2} + \frac{\log{\left(u + 1 \right)}}{2}

        The result is: u36u24\frac{u^{3}}{6} - \frac{u^{2}}{4}

    Now substitute uu back in:

    x2+(2x+1)32614- \frac{x}{2} + \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} - \frac{1}{4}

  2. Now simplify:

    x2+(2x+1)32614- \frac{x}{2} + \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} - \frac{1}{4}

  3. Add the constant of integration:

    x2+(2x+1)32614+constant- \frac{x}{2} + \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} - \frac{1}{4}+ \mathrm{constant}


The answer is:

x2+(2x+1)32614+constant- \frac{x}{2} + \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} - \frac{1}{4}+ \mathrm{constant}

The answer (Indefinite) [src]
  /                                               
 |                                             3/2
 |        x               1       x   (2*x + 1)   
 | --------------- dx = - - + C - - + ------------
 |   _________            4       2        6      
 | \/ 2*x + 1  + 1                                
 |                                                
/                                                 
x2x+1+1dx=Cx2+(2x+1)32614\int \frac{x}{\sqrt{2 x + 1} + 1}\, dx = C - \frac{x}{2} + \frac{\left(2 x + 1\right)^{\frac{3}{2}}}{6} - \frac{1}{4}
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.5
The answer [src]
        ___
  2   \/ 3 
- - + -----
  3     2  
23+32- \frac{2}{3} + \frac{\sqrt{3}}{2}
=
=
        ___
  2   \/ 3 
- - + -----
  3     2  
23+32- \frac{2}{3} + \frac{\sqrt{3}}{2}
-2/3 + sqrt(3)/2
Numerical answer [src]
0.199358737117772
0.199358737117772
The graph
Integral of x/(sqrt(2x+1)+1) dx

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