Neat evaluation of - $\int_1^e\frac{\ln^3x}{(x^2)}dx$ 
Evaluation of - $$\int_1^e\frac{\ln^3x}{(x^2)}dx$$

I've been trying to find a more neat evaluation of this integral without ending up with many separate integrations by paths. Any ideas ? 
 A: Let $\ln x= y\implies x=e^y,dx=e^y\ dy$
$$\int_1^e\dfrac{(\ln x)^3}{x^2}dx=\int_0^1\dfrac{y^3}{e^{2y}}e^y\ dy=\int_0^1e^{-y}y^3\ dy$$
Method $\#1:$
Now integrate by parts 
See LIATE
Method $\#2:$
Or use reduction formula
$$\dfrac{d(e^{ay}y^n)}{dy}=ae^{ay}y^n+ne^{ay}y^{n-1}$$
Integrate both sides to get
$$aI_n+nI_{n-1}=e^{ay}y^n$$ where $I_n=\int e^{ay}y^n\ dy,$
A: For any polynomial $p(x)$ we have
$$ \int p(x) e^{-x}\,dx = C-\left[p(x)+p'(x)+p''(x)+\ldots\right]e^{-x}$$
hence
$$ \int_{1}^{e}\frac{\log^3 z}{z^2}\,dz = \int_{0}^{1}x^3 e^{-x}\,dx = -\left[\left(x^3+3x^2+6x+6\right)e^{-x}\right]_{0}^{1}=6-\frac{16}{e}.$$
A: An effective way to streamline the process of applying integration by parts many times is to use what is known as tabular integration by parts which is nothing more than an efficient way of organising the mechanics of applying integration by parts multiple times. 
As its name suggests, a table is used. It consists of two columns that one usually called $D$ for differentiate (the function to be differentiated when using integration by parts) and $I$ for integrate (the function to be integrated when using integration by parts). The entries in the first row are the functions selected to be differentiated $u$, and integrated $dv$. Each successive entry in the $D$ column is the derivative of its previous column entry while each successive entry in the $I$ column is the integral of its previous column entry. 
The table can be terminated at any time, and when it is, the integral is written down as the sum of the products of the function in the $n$th row of column $D$ with the function in the $(n + 1)$th row of column $I$, with the sign of the products alternating after stating with a plus sign. To help remember the technique the appropriate products between functions are formed by drawing a diagonal arrow in the table while its corresponding sign is placed just above each arrow.
So for your particular integral after a substitution of $x \mapsto e^x$ has been enforced, as
$$I = \int_0^1 x^3 e^{-x} \, dx,$$
if we chose to differentiate $x^3$ and integrate $e^{-x}$ we have:
\begin{array}{|c|c|c|}\hline
D & I & \\\hline
x^3 & e^{-x} & \\
3x^2 & -e^{-x} & +\\
6x & e^{-x} & -\\
6 & -e^{-x} & +\\
0 & e^{-x} & -\\\hline
\end{array}
So on application of the tabular method we have
$$\int_0^1 x^3 e^{-x} \, dx = (-x^3 - 3x^2 - 6x - 6) e^{-x} \Big{|}_0^1 = 6 - \frac{16}{e}.$$
A: For yet another approach, and one that completely avoids integration by parts altogether, Feynman's trick of differentiating under the integral sign can be used.
Let
$$I(a) = \int_1^e x^a \, dx, \quad a \neq -1.$$
Differentiating with respect to the parameter $a$ three times one has
$$I'''(a) = \int_1^e x^a \ln^3 x \, dx,$$
and our integral is recovered on setting $a = -2$.
Since
$$I(a) = \int_1^e x^a \, dx = \left [\frac{x^{a + 1}}{a + 1} \right ]_1^e = \frac{e^{a + 1} - 1}{a + 1},$$
our problem is converted into a problem involving differentiation. Differentiating once, twice, and three times we have
$$I'(a) = \frac{a e^{a + 1} + 1}{(a + 1)^2}, \quad I''(a) = \frac{(a^2 + 1) e^{a + 1} - 2}{(a + 1)^3}, \quad I'''(a) = \frac{(a^3 + 3a - 2) e^{a + 1} + 6}{(a + 1)^4}.$$
On setting $a = -2$ in the third derivative we find
$$I'''(-2) = \int_1^e \frac{\ln^3 x}{x^2} \, dx = 6 - \frac{16}{e},$$
as expected.
