$\displaystyle \int \dfrac{\mathrm{d}x}{\sqrt{x-1}+\sqrt{x+1}}$

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

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

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

$\displaystyle \int \dfrac{\mathrm{d}x}{\sqrt{2x-x^2}}$

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

$\displaystyle \int \tan^3 x\,\mathrm{d}x$

Keeping in mind that $\displaystyle \int \tan x\,\mathrm{d}x=-\ln|\cos x|-k$, which was calculated in a previous problem, we do a trick $\displaystyle \int \tan^3 x\,\mathrm{d}x= \int \tan x(1+ \tan^2 x)\,\mathrm{d}x-\int \tan x\,\mathrm{d}x \,\overset{\substack{t=\tan x\\ \mathrm{d}t=(1+ \tan^2 x)\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int t\,\mathrm{d}t-(-\ln|\cos x|)+k=\\ \qquad\displaystyle \dfrac{1}{2}t^2+\ln|\cos x|+k \,\overset{\substack{t=\tan x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{2}(\tan x)^2+\ln|\cos x|+k}$

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

$\displaystyle \int \dfrac{x^3+3x}{x^4+1}\,\mathrm{d}x= \dfrac{1}{4}\int \dfrac{4x^3}{x^4+1}\,\mathrm{d}x+\dfrac{3}{2}\int \dfrac{2x}{x^4+1}\,\mathrm{d}x \,\overset{\substack{t=x^4\\ \mathrm{d}t=4x^3\,\mathrm{d}x\\w=x^2\\ \mathrm{d}w=2x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{4}\int \dfrac{1}{t+1}\,\mathrm{d}t+\dfrac{3}{2}\int \dfrac{1}{1+w^2}\,\mathrm{d}x=\\ \qquad\displaystyle \dfrac{1}{4}\ln|t+1| + \dfrac{3}{2}\tan^{-1}w + k \,\overset{\substack{t=x^4\\w=x^2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{4}\ln|x^4+1| + \dfrac{3}{2}\tan^{-1}(x^2) + k}$