$\displaystyle \int \dfrac{x^2}{8+x^6}\,\mathrm{d}x$

$\displaystyle \int \dfrac{x^2}{8+x^6}\,\mathrm{d}x= \dfrac{1}{3}\int \dfrac{3x^2}{8+x^6}\,\mathrm{d}x \,\overset{\substack{t=x^3\\ \mathrm{d}t=3x^2\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{3}\int \dfrac{1}{8+t^2}\,\mathrm{d}t= \dfrac{1}{3\sqrt{8}}\tan^{-1}\left(\dfrac{t}{\sqrt{8}}\right) + k \,\overset{\substack{t=x^3\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{3\sqrt{8}}\tan^{-1}\left(\dfrac{x^3}{\sqrt{8}}\right) + k}$

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

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

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

$\displaystyle \int \dfrac{x+4}{\sqrt{1-x^2}}\,\mathrm{d}x= \int \dfrac{x}{\sqrt{1-x^2}}\,\mathrm{d}x + 4\int \dfrac{1}{\sqrt{1-x^2}}\,\mathrm{d}x= -\dfrac{1}{2}\int \dfrac{-2x}{\sqrt{1-x^2}}\,\mathrm{d}x + 4\sin^{-1}x+k \,\overset{\substack{t=1-x^2\\ \mathrm{d}t=-2x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle -\dfrac{1}{2}\int \dfrac{1}{\sqrt{t}}\,\mathrm{d}t + 4\sin^{-1}x+k= -\sqrt{t} + 4\sin^{-1}x + k \,\overset{\substack{t=1-x^2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{-\sqrt{1-x^2} + 4\sin^{-1}x + k}$

$\displaystyle \int \dfrac{1}{x\ln x}\,\mathrm{d}x$

$\displaystyle \int \dfrac{1}{x\ln x}\,\mathrm{d}x \,\overset{\substack{t=\ln x\\ \mathrm{d}t=\frac{1}{x}\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int \dfrac{1}{t}\,\mathrm{d}t= \ln |t|+k \,\overset{\substack{t=\ln x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\ln|\ln x|+k}$

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

$\displaystyle \int \dfrac{x}{\sqrt{1-x^4}}\,\mathrm{d}x= \dfrac{1}{2}\int \dfrac{2x}{\sqrt{1-x^4}}\,\mathrm{d}x \,\overset{\substack{t=x^2\\ \mathrm{d}t=2x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\int \dfrac{1}{\sqrt{1-t^2}}\,\mathrm{d}t= \dfrac{1}{2}\sin^{-1} t + k \,\overset{\substack{t=x^2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{2}\sin^{-1} (x^2) + k}$