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

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

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

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

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

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

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

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

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

$\displaystyle \int \dfrac{3x+2}{4+x^2}\,\mathrm{d}x= 3\int \dfrac{x}{4+x^2}\,\mathrm{d}x+2\int \dfrac{1}{4+x^2}\,\mathrm{d}x$ Let's go with the first part: $\displaystyle \int \dfrac{x}{4+x^2}\,\mathrm{d}x= \dfrac{1}{2}\int \dfrac{2x}{4+x^2}\,\mathrm{d}x \,\overset{\substack{t=4+x^2\\ \mathrm{d}t=2x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\int \dfrac{1}{t}\,\mathrm{d}t= \dfrac{1}{2}\ln|t| + k \,\overset{\substack{t=4+x^2\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\ln|4+x^2| + k = \dfrac{1}{2}\ln(4+x^2) + k$ As for the second part, we know from a previous exercise that: $\displaystyle \dfrac{1}{4+x^2}\,\mathrm{d}x= \dfrac{1}{2}\tan^{-1}\left(\dfrac{x}{2}\right) + k$ So adding everything up we get $\displaystyle \int \dfrac{3x+2}{4+x^2}\,\mathrm{d}x= \bbox[#FFECB3,5px]{\dfrac{3}{2}\ln(4+x^2) + \tan^{-1}\left(\dfrac{x}{2}\right) + k}$