$\partial_x \left( \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt \right)$, somehow skipping the integral? 
Original problem (third problem here) 
Plane $T$ nears the surface $S$ $$S: \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt = \ln(2)$$ in a point that is  on positive $z$ -axis. Assign $T$'s equation.

So I think I need $\nabla S\times T=0$ where $T=\begin{pmatrix}x \\ y \\ z \end{pmatrix}$. So
$$\partial_x S := \partial_x \left( \int_0 ^1 \frac{e^{xzt}}{x-y+z+t} dt \right)=?$$
Now should I integrate this before differentiation or was there some rule to make this differentation-integration simpler?

Trial 4 
  By Leibniz rule kindly suggested by the engineer and the formula by Petersen,
$$\begin{align}\partial_x \int_0^1 f(x,t) dt \Bigg\vert_{x=0} 
&= \int_0^1 \partial_x f(x,t) \Big\vert_{x=0} dt \\
&=z(y-z)\log\left(\frac{1-y+z}{-y+z}\right)+z+\frac{1}{-y+z+1}-\frac{1}{-y+z} \\
&:=B
\end{align}$$
Details here. Now after this, the same for $\partial_y S:=C$ and $\partial_z S :=D$ so $\nabla S = \begin{pmatrix} B \\ C \\ D\end{pmatrix}$. 

But to the question, is $\nabla S = \begin{pmatrix} B \\ ... \\ ...\end{pmatrix}$ right? Or $\nabla S_{|\bar{x}=(0,0,1)} = \begin{pmatrix} -\log(2)+0.5 \\ ... \\ ...\end{pmatrix}$? Look after this I need to do $\nabla S\times T=0$ and I want to minimize the amount of terms early on because it is easy to do mistakes with long monotonous calculations.

Old trials in the chronological order
  
  
*
  
*T0: Hard (or/and indefinite) integral-differential
  
*T1: Leibniz -hint but still indefinite
  
*T2: Plane and surface meeting at the point (0,0,1), deduction with "extra-minus" mistake
  
*T3: trying to calculate the $\nabla$ (this uses wrong point (0,0,-2) instead of (0,0,1) but is the idea correct?) (here)

 A: Gradients of scalar functions are normal aka perpendicular to level surfaces. Planes in $\mathbb{R}^3$ are given by equations of the form $\mathbf{n}\cdot(\mathbf{x}-\mathbf{p})=0$, where $\mathbf{n}$ is a normal vector and $\mathbf{p}$ is a point on the plane.
Thus the plane you're looking for is given by
$$F(x,y,z)=\int_0^1 \frac{\exp(xzt)}{x-y+z+t}dt, \quad  z_0: F(0,0,z_0)=\log2, \quad \nabla F(0,0,z_0)\cdot\big(\vec{x}-(0,0,z_0)\big)$$
The components of $\nabla F$ should all evaluate to concrete quantities. For example,
$$\int_0^1 \frac{\partial}{\partial x}\frac{\exp(xzt)}{x-y+z+t}dt=\int_0^1\left(zt-\frac{1}{x-y+z+t}\right)\frac{\exp(xzt)}{x-y+z+t}dt $$
which, after evaluation, becomes
$$\int_0^1\frac{z_0t}{z_0+t}-\frac{1}{(z_0+t)^2}dt =z_0(1-z_0\log2)-\left(\frac{1}{z_0+1}-\frac{1}{z_0}\right).$$
Also note that $z_0$ solves explicitly to $1/(e^2-1)$.
A: Wolfram Alpha indicates that this integral can not be expressed by elementary functions. This is a big wrong way sign saying that you should try to solve the problem in some other way.
In your problem, you are given that the plane $T$ is tangent to the surface $S$ at a point on the $z$-axis, i.e. a point for which $x=y=0$. Start by finding this point, by solving
$$ \int_0^1 \frac 1{z+t} dt = \ln 2.$$
Then find the partial derivatives at this point. Use that
$$ \partial_x \int_0^1 f(x,t) dt \Bigg\vert_{x=0} = \int_0^1 \partial_x f(x,t) \Big\vert_{x=0} dt.$$
