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

First we perform polynomial division until the degree of the numerator is less than the degree of the denominator

$$ \begin{array}{rcl} x^3 & = & x^3+2x^2+3x-2x^2-3x\\ & = & x(x^2+2x+3)-2x^2-3x\\ & = & x(x^2+2x+3)-2x^2-4x-6+x+6\\ & = & x(x^2+2x+3)-2(x^2+2x+3)+x+6\\ & = & (x-2)(x^2+2x+3)+x+6 \end{array}$$ $$\dfrac{x^3}{x^2+2x+3}=x-2+\dfrac{x+6}{x^2+2x+3}$$ Now we focus on partial fractions decomposition. $x^2+2x+3$ has no real roots

$$x^2+2x+3=0\Longrightarrow x=\dfrac{-2\pm\sqrt{2^2-4\cdot 1\cdot 3}}{2}=\dfrac{-2\pm\sqrt{-8}}{2}$$

Indeed,

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

Therefore

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

$\displaystyle \int \dfrac{x^5}{(x+1)^3}\,\mathrm{d}x$

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

$$\dfrac{x^5}{(x+1)^3}=\dfrac{(y-1)^5}{y^3}=\dfrac{y^5-5y^4+10y^3-10y^2+5y-1}{y^3}=y^2-5y+10-\dfrac{10}{y}+\dfrac{5}{y^2}-\dfrac{1}{y^3}=$$ $$(x+1)^2-5(x+1)+10-\dfrac{10}{x+1}+\dfrac{5}{(x+1)^2}-\dfrac{1}{(x+1)^3}$$

Therefore

$\displaystyle \int \dfrac{x^5}{(x+1)^3}\,\mathrm{d}x= \int \left((x+1)^2-5(x+1)+10-\dfrac{10}{x+1}+\dfrac{5}{(x+1)^2}-\dfrac{1}{(x+1)^3}\right)\,\mathrm{d}x=\\ \qquad\displaystyle \bbox[#FFECB3,5px]{\dfrac{1}{3}(x+1)^3 - \dfrac{5}{2}(x+1)^2+10x -10\ln|x+1| - \dfrac{5}{x+1} + \dfrac{1}{2(x+1)^2} + k}$

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

$\displaystyle \int \dfrac{\mathrm{d}x}{1+e^x}$

$\displaystyle \int \dfrac{\mathrm{d}x}{1+e^x}= \int \dfrac{e^x}{e^x(1+e^x)}\,\mathrm{d}x \,\overset{\substack{t=e^x\\ \mathrm{d}t=e^x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int \dfrac{1}{t(1+t)}\,\mathrm{d}t= \int \left(\dfrac{1}{t}-\dfrac{1}{t+1}\right)\,\mathrm{d}t= \ln|t|-\ln|t+1| + k \,\overset{\substack{t=e^x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{x - \ln(1+e^x) + k}$

$\displaystyle \int \dfrac{5}{x^2-3}\,\mathrm{d}x$

Let's decompose in partial fractions:

$$x^2-3=(x-\sqrt{3})(x+\sqrt{3})$$ $$\dfrac{1}{x-\sqrt{3}}-\dfrac{1}{x+\sqrt{3}}=\dfrac{x+\sqrt{3}}{(x-\sqrt{3})(x+\sqrt{3})}-\dfrac{x-\sqrt{3}}{(x-\sqrt{3})(x+\sqrt{3})}=\dfrac{2\sqrt{3}}{(x-\sqrt{3})(x+\sqrt{3})}$$ $$\dfrac{5}{x^2-3}=\dfrac{5}{2\sqrt{3}}\left(\dfrac{1}{x-\sqrt{3}}-\dfrac{1}{x+\sqrt{3}}\right)$$

Therefore

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

$\displaystyle \int \sqrt{1+e^x}\,\mathrm{d}x$

$\displaystyle \int \sqrt{1+e^x}\,\mathrm{d}x= \int \dfrac{e^x\sqrt{1+e^x}}{e^x}\,\mathrm{d}x \,\overset{\substack{t=e^x\\ \mathrm{d}t=e^x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int \dfrac{\sqrt{t+1}}{t}\,\mathrm{d}t \,\overset{\substack{w=\sqrt{t+1}\\ w^2=t+1 \\ 2w\,\mathrm{d}w=\mathrm{d}t\\ \phantom{\downarrow}}}{=}\, \int \dfrac{2w^2}{w^2-1}\,\mathrm{d}w= \int \left(2+\dfrac{2}{w^2-1}\right)\,\mathrm{d}w=\\ \qquad\displaystyle \int \left(2+\dfrac{1}{w-1}-\dfrac{1}{w-1}\right)\,\mathrm{d}w= 2w+\ln|w-1|-\ln|w+1| + k \,\overset{\substack{w=\sqrt{t+1}\\ \phantom{\downarrow}}}{=}\, 2\sqrt{t+1}+\ln(\sqrt{t+1}-1)-\ln(\sqrt{t+1}+1) + k \,\overset{\substack{t=e^x\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle \bbox[#FFECB3,5px]{2\sqrt{1+e^x}+\ln(\sqrt{1+e^x}-1)-\ln(\sqrt{1+e^x}+1) + k }$