Substitution Method Problem (Calc I)

I would do it like this:

.

Compute the definite integral:
 integral_1^(e^3) x (log(x)+1) dx
Expanding the integrand x (log(x)+1) gives x+x log(x):
  =   integral_1^(e^3) (x+x log(x)) dx
Integrate the sum term by term:
  =   integral_1^(e^3) x log(x) dx+ integral_1^(e^3) x dx
For the integrand x log(x), integrate by parts,  integral f dg = f g- integral g df, where
 f = log(x),     dg = x  dx,
 df = 1/x  dx,     g = x^2/2:
  =  1/2 x^2 log(x)|_1^(e^3)+1/2 integral_1^(e^3) x dx
Evaluate the antiderivative at the limits and subtract.
 1/2 x^2 log(x)|_1^(e^3) = 1/2 (e^3)^2 log(e^3)-1/2 1^2 log(1) = (3 e^6)/2:
  =  (3 e^6)/2+1/2 integral_1^(e^3) x dx
Apply the fundamental theorem of calculus.
The antiderivative of x is x^2/2:
  =  (3 e^6)/2+x^2/4|_1^(e^3)
Evaluate the antiderivative at the limits and subtract.
 x^2/4|_1^(e^3) = (e^3)^2/4-1^2/4 = 1/4 (e^6-1):
  =  (3 e^6)/2+1/4 (e^6-1)
Which is equal to:
Answer: |   =  1/4 (7 e^6-1)

So, not entire sure how to go about doing this one using the substitution method, so i'd appreciate any help :\

$\int_{1}^{e^3} \frac{dx}{x \cdot [1+\ln{(x)} ] }$

$\begin{array}{rcll} \int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } \\\\ && \text{substitute }~z = 1+\ln{(x)} \qquad dz = \frac{dx}{x} \qquad dx = x\cdot dz \\\\ \int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } &=& \int_{1}^{e^3} \dfrac{x\cdot dz}{x \cdot z } \\\\ &=& \int_{1}^{e^3} \dfrac{dz}{ z } \\\\ &=& \left[~ \ln{(z)} ~\right]_{1}^{e^3} \\\\ &=& \left[~ \ln{(~ 1+\ln{(x)} ~)} ~\right]_{1}^{e^3} \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - \ln{(~ 1+\ln{(1)}~)} \qquad | \qquad \ln{(1)} = 0 \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - \ln{( 1+0)} \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - \ln{(1)} \qquad | \qquad \ln{(1)} = 0 \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} - 0 \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} \\\\ &=& \ln{(~ 1+\ln{(e^3)} ~)} \qquad | \qquad \ln{(e^3)} = 3\\\\ &=& \ln{( 1+3 )} \\\\ \int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } &=& \ln{( 4 )} \\\\ \mathbf{\int_{1}^{e^3} \dfrac{dx}{x \cdot [1+\ln{(x)} ] } }&\mathbf{=} & \mathbf{1.38629436112} \\\\ \end{array}$