$\displaystyle \int \dfrac{\mathrm{d}x}{x^4-1}$

Let's decompose in partial fractions:

$$\dfrac{1}{x^4-1}=\dfrac{A}{x+1}+\dfrac{B}{x-1}+\dfrac{C+Dx}{x^2+1}$$ $$1 = A(x-1)(x^2+1)+B(x+1)(x^2+1)+(C+Dx)(x-1)(x+1)$$ $$x=-1\leadsto 1=A(-1-1)((-1)^2+1)+B(-1+1)((-1)^2+1)+(C+D\cdot(-1))(-1-1)(-1+1)=-4A\Longrightarrow A=-\dfrac{1}{4}$$ $$x=1\leadsto 1=A(1-1)(1^2+1)+B(1+1)(1^2+1)+(C+D\cdot 1)(1-1)(1+1)=4B\Longrightarrow B=\dfrac{1}{4}$$ $$x=0\leadsto 1=A(0-1)(0^2+1)+B(0+1)(0^2+1)+(C+D\cdot 0)(0-1)(0+1)=-A+B-C=\dfrac{1}{2}-C\Longrightarrow C=-\dfrac{1}{2}$$ $$x=2\leadsto 1=A(2-1)(2^2+1)+B(2+1)(2^2+1)+(C+D\cdot 2)(2-1)(2+1)=5A+15B+3C+6D=1+6D\Longrightarrow D=0$$

Nevertheless, it may be helpful to keep in mind that

$$\dfrac{1}{a-1}-\dfrac{1}{a+1}=\dfrac{a+1}{a^2-1}-\dfrac{a-1}{a^2-1}=\dfrac{2}{a^2-1}$$ $$\dfrac{1}{a^2-1}=\dfrac{1}{2}\left(\dfrac{1}{a-1}-\dfrac{1}{a+1}\right)$$

which would have made our decomposition much easier

$$\dfrac{1}{x^4-1}=\dfrac{1}{2(x^2-1)}-\dfrac{1}{2(x^2+1)}=\dfrac{1}{4(x-1)}-\dfrac{1}{4(x+1)}-\dfrac{1}{2(x^2+1)}$$ Therefore

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

$\displaystyle \int \dfrac{4x+1}{x^2+x+3}\,\mathrm{d}x$

$x^2+x+3$ has no real roots

$$x^2+x+3=0\Longrightarrow x=\dfrac{-1\pm\sqrt{1^2-4\cdot 1\cdot 3}}{2}=\dfrac{-1\pm\sqrt{-11}}{2}$$

Indeed,

$$\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}$$ $$x^2+x+3=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}$$ $$4x+1=4\left(x+\dfrac{1}{2}\right)-1$$ $$\dfrac{4x+1}{x^2+x+3}=\dfrac{4\left(x+\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)^2+\frac{11}{4}}-\dfrac{1}{\left(x+\frac{1}{2}\right)^2+\frac{11}{4}}$$

Therefore

$\displaystyle \int \dfrac{4x+1}{x^2+x+3}\,\mathrm{d}x= \int \left(\dfrac{4\left(x+\frac{1}{2}\right)}{\left(x+\frac{1}{2}\right)^2+\frac{11}{4}}-\dfrac{1}{\left(x+\frac{1}{2}\right)^2+\frac{11}{4}}\right)\,\mathrm{d}x= 2\ln\left|\left(x+\dfrac{1}{2}\right)^2+\frac{11}{4}\right|-\dfrac{1}{\sqrt{\frac{11}{4}}}\tan^{-1}\left(\dfrac{x+\frac{1}{2}}{\sqrt{\frac{11}{4}}}\right) + k=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{2\ln|x^2+x+3|-\dfrac{2}{\sqrt{11}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{11}}\right) + k}$

$\displaystyle \int \dfrac{2x^2+7x-1}{x^3+x^2-x-1}\,\mathrm{d}x$

Let's decompose in partial fractions:

$$x^3+x^2-x-1=(x+1)(x^2-1)=(x+1)^2(x-1)$$ $$\dfrac{2x^2+7x-1}{x^3+x^2-x-1}=\dfrac{A}{x-1}+\dfrac{B}{x+1}+\dfrac{C}{(x+1)^2}$$ $$2x^2+7x-1 = A(x+1)^2+B(x-1)(x+1)+C(x-1)$$ $$x=-1\leadsto -6=2\cdot (-1)^2+7\cdot (-1)-1 = A(-1+1)^2+B(-1-1)(-1+1)+C(-1-1)=-2C\Longrightarrow C=3$$ $$x=1\leadsto 8=2\cdot 1^2+7\cdot 1-1 = A(1+1)^2+B(1-1)(1+1)+C(1-1)=4A\Longrightarrow A=2$$ $$4x+7=(2x^2+7x-1)'=(A(x+1)^2+B(x-1)(x+1)+C(x-1))'=2A(x+1)+B(x+1)+B(x-1)+C$$ $$x=-1\leadsto 3=4\cdot(-1)+7=2A(-1+1)+B(-1+1)+B(-1-1)+C=-2B+3\Longrightarrow B=0$$

Therefore

$\displaystyle \int \dfrac{2x^2+7x-1}{x^3+x^2-x-1}\,\mathrm{d}x= \int \left(\dfrac{2}{x-1}+\dfrac{3}{(x+1)^2}\right)\,\mathrm{d}x= \bbox[#FFECB3,5px]{2\ln|x-1|-\dfrac{3}{x+1} + k}$

$\displaystyle \int \dfrac{3x+2}{x^3+4x^2+4x}\,\mathrm{d}x$

Let's decompose in partial fractions:

$$x^3+4x^2+4x=x(x^2+4x+4)=x(x+2)^2$$ $$\dfrac{3x+2}{x^3+4x^2+4x}=\dfrac{A}{x}+\dfrac{B}{x+2}+\dfrac{C}{(x+2)^2}$$ $$3x+2 = A(x+2)^2+Bx(x+2)+Cx$$ $$x=-2\leadsto -4=3\cdot (-2)+2 = A(-2+2)^2+B(-2)(-2+2)+C(-2)=-2C\Longrightarrow C=2$$ $$x=0\leadsto 2=3\cdot 0+2=A(0+2)^2+B\cdot 0(0+2)+C\cdot 0=4A\Longrightarrow A=\dfrac{1}{2}$$ $$3=(3x+2)'=(A(x+2)^2+Bx(x+2)+Cx)'=2A(x+2)+B(x+2)+Bx+C$$ $$x=-2\leadsto 3=2A(-2+2)+B(-2+2)+B\cdot (-2)+C=-2B+2\Longrightarrow B=-\dfrac{1}{2}$$

Therefore

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

$\displaystyle \int \dfrac{3x+4}{x^2+4x+3}\,\mathrm{d}x$

Let's decompose in partial fractions:

$$x^2+4x+3=(x+1)(x+3)$$ $$\dfrac{3x+4}{x^2+4x+3}=\dfrac{A}{x+1}+\dfrac{B}{x+3}$$ $$3x+4 = A(x+3)+B(x+1)$$ $$x=-3\leadsto -5=3\cdot (-3)+4 = A(-3+3)+B(-3+1)=-2B\Longrightarrow B=\dfrac{5}{2}$$ $$x=-1\leadsto 1=3\cdot (-1)+4 = A(-1+3)+B(-1+1)=2A\Longrightarrow A=\dfrac{1}{2}$$

Therefore

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