Tough definite integral I had this integral in my assignment and I can't figure out a way to solve this. I can see that in the bracket, there is $x \ln x$ and $\ln(x) +1$ which is the derivative of $x \ln x$ so I thought maybe some substitution; but the exponential powers of $e$ destroyed that hope. How do I solve this integral?

Let$$
I_n = \int_1^{1+\frac{1}{n}}\left\{[(x+1)\ln x +1] e^{x (e^x \ln x +1)}+n \right\}\, dx\qquad (n=1,2,\ldots)
$$
  Evaluate $ \lim_{n \to \infty} I_n^n$.

 A: This is what seems probable. Substituting $u=x\exp(x)\ln(x)$, so that $\mathrm du=\exp(x)((x+1)\ln(x)+1)\mathrm dx$. Not sure though how to get rid of that extra $\exp(x)$ in there. Or better yet, substitute $u=\exp(x)$. 

On solving the antiderivative comes out to be $x^{x\exp(x)}$ and $I_n^{n}$ comes out to be the following expression and taking limit as $n\to \infty$, we get 
$$\lim_{n\to \infty}I_n^{n}=\lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n(1+1/n)\exp(1+1/n)}=\exp\left(\lim_{n\to \infty}\frac{(n+1)\exp\left((1+\frac{1}{n}\right))}{n}\right)=\exp(e)$$
A: One thing you could do is to compose a Taylor expansion of the integrand around $x=1$. This should give
$$\big[(x+1)\log(x) +1\big]\, e^{x (e^x \log(x) +1)}=e+e\left(3 +e\right) (x-1)+O\left((x-1)^2\right)$$ Using the above approximation
$$\int_1^{1+\frac 1n}\big[e+e\left(3 +e\right) (x-1)+n\big]\,dx=1+\frac{e}{n}+\frac{e(3 +e)}{2 n^2}+O\left(\frac{1}{n^3}\right)$$
Comparing with numerical integration
$$\left(
\begin{array}{ccc}
n & \text{approximation} & \text{exact} \\
 10 & 1.34955 & 1.37021 \\
 20 & 1.15534 & 1.15766 \\
 30 & 1.09924 & 1.09991 \\
 40 & 1.07281 & 1.07309 \\
 50 & 1.05747 & 1.05761 \\
 60 & 1.04746 & 1.04754 \\
 70 & 1.04042 & 1.04047 \\
 80 & 1.03519 & 1.03523 \\
 90 & 1.03116 & 1.03119 \\
 100 & 1.02796 & 1.02798
\end{array}
\right)$$
Edit
Since @Paras Khosla  did show it, the antiderivative is "just"
$$x^{x\,e^x }+n x$$ making the integral to be
$$I_n=\left(1+\frac{1}{n}\right)^{\frac{ (n+1)}{n}e^{1+\frac{1}{n}}}$$ Using Taylor series for large $n$ leads to the same as above.
A: At the ${ e }^{ x({ e }^{ x }(\ln { x } +1) }$,
notice the +1. It means an ${ e }^{ x}$ can get out at front.
Then, it becomes $$ \{ { e }^{ x }x\ln { x } +{ e }^{ x }\ln { x } +{ e }^{ x }\} { e }^{ x{ e }^{ x }\ln{x}} $$
Integrate the given formula, and it becomes
$$\lim _{ n\rightarrow \infty  }{ { e }^{ (n+1){ e }^{ (1+\frac { 1 }{ n } ) }\ln { (1+\frac { 1 }{ n } ) }  }}$$
then(after a brief calculation) ,
$$ \quad { e }^{ e\lim _{ n\rightarrow \infty  }{ \frac { \ln { (1+\frac { 1 }{ n } ) }  }{ \frac { 1 }{ n+1 }  }  }  }$$
So the limit is $ { e }^{ e } $.
A: Let $$f(x) =[(x+1)\log x+1]e^{x(e^x\log x +1)}$$ then we have $$I_n=\int_{1}^{1+(1/n)}\{f(x)+n\}\,dx=1+\int_{1}^{1+(1/n)}f(x)\,dx=1+g(n)$$ where clearly $g(n) \to 0$ as $n\to\infty $.
Now $$I_n^n=\exp(n\log I_n) =\exp\left(\frac{\log(1+g(n))}{g(n)}\cdot ng(n) \right) $$ By fundamental theorem of calculus $ng(n) \to f(1)=e$ as $n\to\infty $ and thus the desired limit is $$\exp(1\cdot e) =e^e$$ There is no need to use anti-derivatives here. 
A: Firstly,
let $t =  e^{x (e^x \ln x +1)}$
Find $dt/dx$.
You will observe that by finding $dt/dx$ You will find something in product. Then you apply integration by parts.
