We will show that the space contains isomorphically the space $l_1$ therefore the space  is not reflexive. We start with the following that I  posed as a question in a previous comment.  
    
  **Fact 1:**  For every $0<\delta < 1 $ $lim_n \frac{\int_1^{1+\delta} n^{1/p} dp } {\int_1^2  n^{1/p} dp } = 1$.  
  
**Proof:**  Since the function $n^{1/p}$, $1\leq p \leq 2$ is decreasing we have that  $\int_1^{1+\delta} n^{1/p} dp >\delta n^{1/1+\delta}$ and  
 $n^ {1/1+2\delta} > \int_{1+2\delta}^2 n^{1/p} dp$   
 Now for $n\in N$ we have that $ \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2  n^{1/p} dp } <   \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^{1+\delta}  n^{1/p} dp } < \frac{n^{1/1+2\delta }} {\delta n^{1/1+ \delta }} = \frac{1} { \delta} \frac {1}{n^{\delta/ (1+\delta)(1+2\delta)}}$.  
 Hence  for every $0<\delta <1/2$  
 
$lim_n  \frac{\int_{1+2\delta}^2 n^{1/p}dp} {\int_1^2  n^{1/p} dp } =0$ which finishes the proof of Fact 1.  
  
 **Fact 2:**  We start with the following classical result.  
If $(f_n)_n$ is a normalized sequence in $L^1 [1,2]$ which is not uniformly integrable (i.e. there exists $\epsilon>0$ such that for every $\delta>0$ 
there exists a Borel set $A$ with $\lambda (A)<\delta$ and $\int_A |f_n| >  \epsilon$ for infinite  $n\in N$ ) then $(f_n)_n$ has a subsequnce equivalent to $l_1$ basis.  
This result is due to Kadec and Pelczynski ( see **J. Diestel: Sequences and Series in Banach Spaces (Graduate Texts in Mathematics, 92) p. 93**).  
Next in the space we denote $(e_i)_i\in N$  the basis of $l_1$ which is a symmetric basis for the space.For $n\in N $ we set $z_n = \sum_ {i=1} ^{n} e_i$ and $x_n = \frac{1} {\int_1^2  n^{1/p} dp } z_n$ which has norm 1.  
 Consider the function $f_n(p) = |x_n|_p $ $1\leq p \leq 2 $ and Fact 1 yields that the sequence $(f_n)$  it is not uniformly integrable.Therefore $(f_n)$ has a subsequence equivalent to $l_1$ basis which implies that $(x_n)$ satisfies the same property in the norm of the space. We will adapt Kadec - Pelczynski's proof in the setting of $(x_n)$.  
 **Step 1 :** There exists a decreasing sequence $(\delta_k )$ and a subsequence $(x_{n_k})$ such that for all $k$ we have that $\int_{1+\delta_{k+1}}^{1+\delta_k} |{x_{n_k}}|_p dp > \frac{1} { 4}$.  
 The proof uses induction and the following :  
From Fact 1 for  $ 0< \delta < 1$ there exists $n\in N$ such that $\int_1^{1+\delta}| {x_n}|_p dp > \frac{1} {2}$. For this $n$ there exists $\delta_1 < \delta $ such that $\int_1^{1+\delta_1}|{x_n}|_p  dp < \frac{1} {4}$. Hence  $\int_{1+\delta_1}^{1+\delta} |{x_n}|_p dp > \frac{1} {4}$.  
  **Step 2:** We set $A_k = [\delta_{k+1}, \delta_{k }]$. There exists an infinite  $I \subset N $  such that for every $k\in I$ setting $ B_k = \cup \{ A_j : j\in I, j\neq k \} $ we have that $\int_{B_k} |{x_{n_k}|_p}dp < \frac {1} {8}$.  
 This is a classical result due to H. P. Rosenthal and an elegant and short proof was given by J. Kupka ( see Page 82 in the aforementioned reference). We assume that $ I = N $ 

  **Claim:** The sequence  $(x_{n_k} )_{k\in N}$ is equivalent to $l_1$ basis.  
  

  
 Indeed   
 $\int_1^2 |\sum_{j=1}^k \lambda_{j} x_{n_j} |_p dp > \int_{\cup_{i\in N} A_i} |\sum_{j=1}^k \lambda_{j} x_{n_j} |_pdp  \geq \sum_{j=1} ^{k} (\int_{A_j} |\lambda_{j} x_{n_j}|_pdp - \int_{B_j} |\lambda_{j} x_{n_j}|_p dp)\geq   
\frac {1}{8} \sum_{j=1}^{k} |\lambda_j|$.

  I have two questions related to this result.  
  
  
  **Question 1 :** Does the space contain a complemented subspace isomorphic to $l_1$ ?    
  
**Edit:** The answer to Question 1 is affirmative hence the dual of the space contains isomorphically the space $l_\infty $. 

  
**Question 2:** Is the space $l_1$ saturated?