Find $$ \int \frac{\sqrt{2+\ln z}}{z} d z $$ Answer: $$ \square $$

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Calculate the integral $$ \int \frac{\sqrt{2+\ln z}}{z} dz $$.
Evaluate the integral by following steps:
- step0: Evaluate using substitution:
  $$\int \frac{\sqrt{2+\ln{\left(zight)}}}{z} dz$$
- step1: Rewrite the expression:
  $$\int \frac{\left(2+\ln{\left(zight)}ight)^{\frac{1}{2}}}{z} dz$$
- step2: Use the substitution $$dz=z dt$$ to transform the integral$$:$$
  $$\int \frac{\left(2+\ln{\left(zight)}ight)^{\frac{1}{2}}}{z}	imes z dt$$
- step3: Simplify:
  $$\int \left(2+\ln{\left(zight)}ight)^{\frac{1}{2}} dt$$
- step4: Use the substitution $$t=\ln{\left(zight)}$$ to transform the integral$$:$$
  $$\int \left(2+tight)^{\frac{1}{2}} dt$$
- step5: Use the substitution $$dt=1 dv$$ to transform the integral$$:$$
  $$\int \left(2+tight)^{\frac{1}{2}}	imes 1 dv$$
- step6: Simplify:
  $$\int \left(2+tight)^{\frac{1}{2}} dv$$
- step7: Use the substitution $$v=2+t$$ to transform the integral$$:$$
  $$\int v^{\frac{1}{2}} dv$$
- step8: Evaluate the integral:
  $$\frac{v^{\frac{1}{2}+1}}{\frac{1}{2}+1}$$
- step9: Add the numbers:
  $$\frac{v^{\frac{3}{2}}}{\frac{1}{2}+1}$$
- step10: Add the numbers:
  $$\frac{v^{\frac{3}{2}}}{\frac{3}{2}}$$
- step11: Substitute back:
  $$\frac{\left(2+tight)^{\frac{3}{2}}}{\frac{3}{2}}$$
- step12: Substitute back:
  $$\frac{\left(2+\ln{\left(zight)}ight)^{\frac{3}{2}}}{\frac{3}{2}}$$
- step13: Simplify:
  $$\frac{\sqrt{\left(2+\ln{\left(zight)}ight)^{3}}}{\frac{3}{2}}$$
- step14: Add the constant of integral C:
  $$\frac{\sqrt{\left(2+\ln{\left(zight)}ight)^{3}}}{\frac{3}{2}} + C, C \in \mathbb{R}$$
The integral $$ \int \frac{\sqrt{2+\ln z}}{z} dz $$ is equal to $$ \frac{\sqrt{(2+\ln{(z)})^{3}}}{\frac{3}{2}} + C $$, where $$ C $$ is an arbitrary constant.
$$ \frac{\sqrt{(2+\ln{(z)})^{3}}}{\frac{3}{2}} + C $$

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