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Question: How do you integrate \[z{\left( {\ln z} \right)^2}dz\] ?...

How do you integrate z(ln⁡z)2dzz{\left( {\ln z} \right)^2}dz ?

Explanation

Solution

Hint : In order to solve this problem, we need to use integration by parts method. So, we will definitely use a method of substitution. We will first put the value of u as (ln⁡z)2{\left( {\ln z} \right)^2} and then we will proceed. Again, we need to use integration by parts so we will substitute u as ln⁡z\ln z . So, let’s begin.
Formulas used:
∫xndx=xn+1n+1+C\int {{x^n}dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C
ddxln⁡x=1x\dfrac{d}{{dx}}\ln x = \dfrac{1}{x}
∫udv=uv−∫vdu\int {udv} = uv - \int {vdu}

Complete step-by-step answer :
Given that,
integrate z(ln⁡z)2dzz{\left( {\ln z} \right)^2}dz
this can be written as
∫z(ln⁡z)2dz\int {z{{\left( {\ln z} \right)}^2}dz}
Now there are two functions in the integration. So we will use integration by parts(IBP) here.
∫udv=uv−∫vdu\int {udv} = uv - \int {vdu}
Put u=(ln⁡z)2u = {\left( {\ln z} \right)^2}
Then differentiating, du=2(ln⁡z)1zdzdu = 2\left( {\ln z} \right)\dfrac{1}{z}dz
dv=zdzdv = zdz
Taking the integral on both sides,
v=z22v = \dfrac{{{z^2}}}{2}
Now putting these values in the integral above we get,
∫z(ln⁡z)2dz=(ln⁡z)2z22−∫z22×2(ln⁡z)1zdz\int {z{{\left( {\ln z} \right)}^2}dz} = {\left( {\ln z} \right)^2}\dfrac{{{z^2}}}{2} - \int {\dfrac{{{z^2}}}{2}} \times 2\left( {\ln z} \right)\dfrac{1}{z}dz
Cancelling z and 2 from the second integral,
=(ln⁡z)2z22−∫z(ln⁡z)dz= {\left( {\ln z} \right)^2}\dfrac{{{z^2}}}{2} - \int z \left( {\ln z} \right)dz
Now again we have two functions in the second integral so we will use IBP again but this time the substitution will be different.
Put u=ln⁡zu = \ln z
Taking the differentiation, we get du=1zdzdu = \dfrac{1}{z}dz
dv=zdzdv = zdz
Taking the integration, we get v=z22v = \dfrac{{{z^2}}}{2}
Now putting these values in the integration above we get,
=(ln⁡z)2z22−[ln⁡z.z22−∫z22×1zdz]= {\left( {\ln z} \right)^2}\dfrac{{{z^2}}}{2} - \left[ {\ln z.\dfrac{{{z^2}}}{2} - \int {\dfrac{{{z^2}}}{2} \times \dfrac{1}{z}} dz} \right]
Again, cancelling z from the second integral,
=(ln⁡z)2z22−[ln⁡z.z22−12∫zdz]= {\left( {\ln z} \right)^2}\dfrac{{{z^2}}}{2} - \left[ {\ln z.\dfrac{{{z^2}}}{2} - \dfrac{1}{2}\int z dz} \right]
Now taking the simple integral using the formula number 1 mentioned above in the list we get,
=(ln⁡z)2z22−[ln⁡z.z22−12z22]+c= {\left( {\ln z} \right)^2}\dfrac{{{z^2}}}{2} - \left[ {\ln z.\dfrac{{{z^2}}}{2} - \dfrac{1}{2}\dfrac{{{z^2}}}{2}} \right] + c
Multiplying the bracket terms by minus sign we get,
=(ln⁡z)2z22−ln⁡z.z22+12z22+c= {\left( {\ln z} \right)^2}\dfrac{{{z^2}}}{2} - \ln z.\dfrac{{{z^2}}}{2} + \dfrac{1}{2}\dfrac{{{z^2}}}{2} + c
Taking the term z22\dfrac{{{z^2}}}{2} common we can write the simplified answer as,
=z22[(ln⁡z)2−ln⁡z.+12]+c= \dfrac{{{z^2}}}{2}\left[ {{{\left( {\ln z} \right)}^2} - \ln z. + \dfrac{1}{2}} \right] + c
This is the correct integral of the given problem.
So, the correct answer is “z22[(ln⁡z)2−ln⁡z.+12]+c\dfrac{{{z^2}}}{2}\left[ {{{\left( {\ln z} \right)}^2} - \ln z. + \dfrac{1}{2}} \right] + c”.

Note : Here we observe that there are two functions given in the integral. So definitely we are going to use integration by parts here. The thing that is to be noted is that logarithmic function is raised to a power so when we take integral, we need to take the integral of power and then the function. So don’t forget. Also note that IBP is to be taken one more time because again we observe two functions in one integral.