(a) identify the transformation, and (b) graphically represent the transform

What's the best method to integrate ${\displaystyle \int \arctan \left\{\frac{1}{1-x}\right\}dx}$?

Evaluate the following integral.
$\int 2x^3+3x-2dx$

Write the parametric equations $x=3t-1,y=3-3t$ as a function of x in the Cartesian form. $y=?$

Evaluate $\int \frac{4}{7}\ln (6t)dt$.

Let $f$ be continuous on $I=[a,b]$ and let $H:I\to \mathbb{R}$ be defined by $H(x)={\int }_x^{b}f(t)dt,x\in I$. Find $H^{\prime }(x)$.

What is the distance between the following polar coordinates?:
${\displaystyle (2,\frac{3\pi }{4}),(1,\frac{15\pi }{8})}$

Approximating $y^{\prime }(t^n)$ at the relation $y^{\prime }(t^n)=f(t^n,y(t^n))$ with the difference quotient $\left[\frac{y(t^{n+1})-y(t^n)}{h}\right]$ we get to the Euler method.
Approximating the same derivative with the quotient $\left[\frac{y(t^n)-y(t^{n-1})}{h}\right]$ we get to the backward Euler method
$y^{n+1}=y^n+hf(t^{n+1},y^{n+1}),n=0,\dots ,N-1$
where $y^0:=y_0$.
In order to find the formula for the forward Euler method, we use the limit $\underset{h\to 0}{lim}\frac{y(x_0+h)-y(x_0)}{h}$ for $x_0=t^n,h=t^{n+1}-t^n$.
In order to find the formula for the backward Euler method, could we pick $h=t^{n-1}-t^n$ although it is negative?
Or how do we get otherwise to the approximation:
$y^{\prime }(t^n)\approx \frac{y(t^n)-y(t^{n-1})}{h}$