$\displaystyle \int \dfrac{5x-1}{x^2-6x+9}\,\mathrm{d}x$

$\displaystyle \int \dfrac{5x-1}{x^2-6x+9}\,\mathrm{d}x= \int \dfrac{5x-15+14}{(x-3)^2}\,\mathrm{d}x= \int \left(\dfrac{5}{x-3}+\dfrac{14}{(x-3)^2}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{5\ln|x-3| -\dfrac{14}{x-3} + k}$

$\displaystyle \int \dfrac{2x^2+x+1}{(x+3)(x-1)^2}\,\mathrm{d}x$

Let's decompose in partial fractions:

$$\dfrac{2x^2+x+1}{(x+3)(x-1)^2}=\dfrac{A}{x+3}+\dfrac{B}{x-1}+\dfrac{C}{(x-1)^2}$$ $$2x^2+x+1 = A(x-1)^2+B(x+3)(x-1)+C(x+3)$$ $$x=-3\leadsto 16=2\cdot(-3)^2+(-3)+1=A(-3-1)^2+B(-3+3)(-3-1)+C(-3+3)=16A\Longrightarrow A=1$$ $$x=1\leadsto 4=2\cdot 1^2+1+1=A(1-1)^2+B(1+3)(1-1)+C(1+3)=4C\Longrightarrow C=1$$

To find $B$, we may isolate it, since we already know $A$ and $C$, but it will be easier to differentiate the expression above once:

$$(2x^2+x+1)' = (A(x-1)^2+B(x+3)(x-1)+C(x+3))'$$ $$4x+1 = 2A(x-1)+B(x-1)+B(x+3)+C$$ $$x=1\leadsto 5=4\cdot 1+1=2A(1-1)+B(1-1)+B(1+3)+C=4B+1\Longrightarrow B=1$$

Therefore

$\displaystyle \int \dfrac{2x^2+x+1}{(x+3)(x-1)^2}\,\mathrm{d}x= \int \left(\dfrac{1}{x+3}+\dfrac{1}{x-1}+\dfrac{1}{(x-1)^2}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{\ln|x+3|+\ln|x-1|-\dfrac{1}{x-1} + k}$

$\displaystyle \int \dfrac{\mathrm{d}x}{x^3-x}$

Let's decompose in partial fractions:

$$\dfrac{1}{x^3-x}=\dfrac{A}{x}+\dfrac{B}{x-1}+\dfrac{C}{x+1}$$ $$1 = A(x-1)(x+1)+Bx(x+1)+Cx(x-1)$$ $$x=0\leadsto 1=A(0-1)(0+1)+B\cdot 0\cdot (0+1)+C\cdot 0\cdot (0-1)=-A\Longrightarrow A=-1$$ $$x=1\leadsto 1=A(1-1)(1+1)+B\cdot 1\cdot (1+1)+C\cdot 1\cdot (1-1)=2B\Longrightarrow B=\dfrac{1}{2}$$ $$x=-1\leadsto 1=A(-1-1)(-1+1)+B\cdot (-1)\cdot (-1+1)+C\cdot (-1)\cdot (-1-1)=2C\Longrightarrow C=\dfrac{1}{2}$$

Therefore

$\displaystyle \int \dfrac{\mathrm{d}x}{x^3-x}= \int \left(\dfrac{-1}{x}+\dfrac{1}{2(x-1)}+\dfrac{1}{2(x+1)}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{-\ln|x|+\dfrac{1}{2}\ln|x-1|+\dfrac{1}{2}\ln|x+1| + k}$

$\displaystyle \int \dfrac{x}{(x+2)^4}\,\mathrm{d}x$

$\displaystyle \int \dfrac{x}{(x+2)^4}\,\mathrm{d}x= \int \dfrac{x+2-2}{(x+2)^4}\,\mathrm{d}x= \int \left(\dfrac{1}{(x+2)^3}-\dfrac{2}{(x+2)^4}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{-\dfrac{1}{2(x+2)^2}+\dfrac{2}{3(x+2)^3} +k}$

$\displaystyle \int \dfrac{3x^2+2}{(x+1)^3}\,\mathrm{d}x$

The partial fraction decomposition will be more easily handled with a change of variable $y=x+1$

$$\dfrac{3x^2+2}{(x+1)^3}=\dfrac{3(y-1)^2+2}{y^3}=\dfrac{3y^2-6y+5}{y^3}=\dfrac{3}{y}-\dfrac{6}{y^2}+\dfrac{5}{y^3}=\dfrac{3}{x+1}-\dfrac{6}{(x+1)^2}+\dfrac{5}{(x+1)^3}$$

Therefore

$\displaystyle \int \dfrac{3x^2+2}{(x+1)^3}\,\mathrm{d}x= \int \left(\dfrac{3}{x+1}-\dfrac{6}{(x+1)^2}+\dfrac{5}{(x+1)^3}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{3\ln|x+1| + \dfrac{6}{x+1} - \dfrac{5}{2(x+1)^2} + k}$

Of course a change of variable $t=x+1$ inside the integral would have had analogous results