1) I got the $A=2$ and $B=-1$. I think I'm just having trouble with the integration part now.
Evaluate the integral $$\int_0^1 \frac{x-6}{x^2-6x+8}\,dx.$$
2) For this one I got that $A=7$, $B=-1$, and $C=1$. I put $\ln |x|^3 - \ln |x+1| + \ln|x-1| + C$ but I keep getting this wrong.
Evaluate the integral. (Remember to use $|u|$ where appropriate. Use $C$ for the constant of integration.) $$\int\frac{7x^2+2x-7}{x^3-x}\,dx$$
3) For this one I have $A=7$, $B=0$, $C=0$, and $D=9$. I put $\ln\bigl(|x^2+1||x^2+5|\bigr)+C$
$$\int\frac{7x^3+9x^2+35x+9}{(x^2+1)(x^2+5)}\,dx$$
4)I used long division and got $x+\frac{-36x+36}{x^2+36}$. Again, I'm having a lot of trouble integrating.
$$\int\frac{x^3+36}{x^2+36}\,dx$$
For $(1)$ as $x^2-6x+8=(x-4)(x-2)$
use Partial Fraction Decomposition to write $$\frac{x-6}{x^2-6x+8}=\frac A{x-4}+\frac B{x-2}$$
Multiply either sides by $x^2-6x+8=(x-4)(x-2),$ to get $x-6=A(x-2)+B(x-4)=x(A+B)-(2A+4B)$
Compare the coefficient of $x$ and the constant to find $A,B$
The link to the solutions of $(2),(4)$ are given in the comment
For $(3),$
Method $1:$ By observation $7x^3+9x^2+35x+9=7x(x^2+5)+9(x^2+1)$
$$\implies \frac{7x^3+9x^2+35x+9}{(x^2+1)(x^2+5)}=\frac{7x(x^2+5)+9(x^2+1)}{(x^2+1)(x^2+5)}=\frac{7x}{x^2+1}+\frac9{x^2+5}$$
For the first part, use $x^2+1=u$ and for the second $\displaystyle\int\frac{dx}{x^2+a^2}=\frac1a\arctan\frac xa$
Method $2:$ Using Partial Fraction Decomposition I think you have already used
$$\frac{7x^3+9x^2+35x+9}{(x^2+1)(x^2+5)}=\frac{Ax+B}{x^2+1}+\frac{Cx+D}{x^2+5}$$
Multiply either sides by $(x^2+1)(x^2+5)$ and compare the coefficients of the different powers namely, $3,2,1,0$ of $x$ to find $A,B,C,D$