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Solving integrals/ dz/z(z^2+4)

Integral of dz/z(z^2+4) dx

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The solution

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    1                 
    /                 
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   |     1 / 2    \   
   |   1*-*\z  + 4/ dz
   |     z            
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-1 + z
∫z−111⋅1z(z2+4) dz\int\limits_{z - 1}^{1} 1 \cdot \frac{1}{z} \left(z^{2} + 4\right)\, dz 
Integral(1*(z^2 + 4)/z, (z, -1 + z, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=z2u = z^{2}.

      Then let du=2zdzdu = 2 z dz and substitute du2\frac{du}{2}:

      ∫u+44u du\int \frac{u + 4}{4 u}\, du

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

        ∫u+42u du=∫u+4u du2\int \frac{u + 4}{2 u}\, du = \frac{\int \frac{u + 4}{u}\, du}{2}

        1. Rewrite the integrand:

          u+4u=1+4u\frac{u + 4}{u} = 1 + \frac{4}{u}

        2. Integrate term-by-term:

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

            ∫1 du=u\int 1\, du = u

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

            ∫4u du=4∫1u du\int \frac{4}{u}\, du = 4 \int \frac{1}{u}\, du

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

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

          The result is: u+4log⁡(u)u + 4 \log{\left(u \right)}

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

      Now substitute uu back in:

      z22+2log⁡(z2)\frac{z^{2}}{2} + 2 \log{\left(z^{2} \right)}

    Method #2

    1. Rewrite the integrand:

      1⋅1z(z2+4)=z+4z1 \cdot \frac{1}{z} \left(z^{2} + 4\right) = z + \frac{4}{z}

    2. Integrate term-by-term:

      1. The integral of znz^{n} is zn+1n+1\frac{z^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫z dz=z22\int z\, dz = \frac{z^{2}}{2}

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

        ∫4z dz=4∫1z dz\int \frac{4}{z}\, dz = 4 \int \frac{1}{z}\, dz

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

        So, the result is: 4log⁡(z)4 \log{\left(z \right)}

      The result is: z22+4log⁡(z)\frac{z^{2}}{2} + 4 \log{\left(z \right)}

  2. Add the constant of integration:

    z22+2log⁡(z2)+constant\frac{z^{2}}{2} + 2 \log{\left(z^{2} \right)}+ \mathrm{constant}

The answer is:

z22+2log⁡(z2)+constant\frac{z^{2}}{2} + 2 \log{\left(z^{2} \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                    
 |                        2            
 |   1 / 2    \          z         / 2\
 | 1*-*\z  + 4/ dz = C + -- + 2*log\z /
 |   z                   2             
 |                                     
/
∫1⋅1z(z2+4) dz=C+z22+2log⁡(z2)\int 1 \cdot \frac{1}{z} \left(z^{2} + 4\right)\, dz = C + \frac{z^{2}}{2} + 2 \log{\left(z^{2} \right)}
Simplify
The answer [src] 
                            2
1                   (-1 + z) 
- - 4*log(-1 + z) - ---------
2                       2
−(z−1)22−4log⁡(z−1)+12- \frac{\left(z - 1\right)^{2}}{2} - 4 \log{\left(z - 1 \right)} + \frac{1}{2} 
=
= 
                            2
1                   (-1 + z) 
- - 4*log(-1 + z) - ---------
2                       2
−(z−1)22−4log⁡(z−1)+12- \frac{\left(z - 1\right)^{2}}{2} - 4 \log{\left(z - 1 \right)} + \frac{1}{2}
Упростить

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

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