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$\displaystyle \int \tan^{-1} x\,\mathrm{d}x$

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

$\displaystyle \int \sin^{-1} x\,\mathrm{d}x$

$\displaystyle \int \sin^{-1} x\,\mathrm{d}x \,\overset{\substack{ u=\sin^{-1} x\quad u'=\frac{1}{\sqrt{1-x^2}}\\ v'=1\quad v=x\\ \phantom{\downarrow}}}{=}\, x\sin^{-1} x-\int \frac{x}{\sqrt{1-x^2}}\,\mathrm{d}x= \bbox[#FFECB3,5px]{x\sin^{-1} x+\sqrt{1-x^2}+k}$

$\displaystyle \int \dfrac{\ln\left(\sqrt{x}\right)}{\sqrt{x}}\,\mathrm{d}x$

$\displaystyle \int \dfrac{\ln\left(\sqrt{x}\right)}{\sqrt{x}}\,\mathrm{d}x \,\overset{\substack{ u=\ln\left(\sqrt{x}\right)\quad u'=\frac{1}{\sqrt{x}}\frac{1}{2\sqrt{x}}\\ v'=\frac{1}{\sqrt{x}}\quad v=2\sqrt{x}\\ \phantom{\downarrow}}}{=}\, 2\sqrt{x}\ln\left(\sqrt{x}\right) - \int\frac{1}{2x}2\sqrt{x}\,\mathrm{d}x= \bbox[#FFECB3,5px]{\sqrt{x}\ln x - 2\sqrt{x} + k}$ Keep in mind that $\ln\left(\sqrt{x}\right)=\dfrac{1}{2}\ln x$. A change of variable $t=\sqrt{x}$ would have worked as well

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

$\displaystyle \int x^3 e^{x^2}\,\mathrm{d}x \,\overset{\substack{t=x^2\\ \mathrm{d}t=2x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\int t e^t\,\mathrm{d}t \,\overset{\substack{ u=t\quad u'=1\\ v'=e^t\quad v=e^t\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}\left(t e^t-\int e^t\,\mathrm{d}t\right)= \dfrac{1}{2}(t-1)e^t + k \,\overset{\substack{t=x^2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{2}(x^2-1)e^{x^2} + k}$

$\displaystyle \int x\tan^{-1} x\,\mathrm{d}x$

$\displaystyle \int x\tan^{-1} x\,\mathrm{d}x \,\overset{\substack{ u=\tan^{-1} x\quad u'=\frac{1}{1+x^2}\\ v'=x\quad v=\frac{1}{2}x^2\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{2}x^2\tan^{-1} x - \int \dfrac{1}{2}x^2\frac{1}{1+x^2}\,\mathrm{d}x= \dfrac{1}{2}x^2\tan^{-1} x - \dfrac{1}{2}\int \left(1-\frac{1}{1+x^2}\right)\,\mathrm{d}x=\\ \qquad\displaystyle \dfrac{1}{2}x^2\tan^{-1} x - \dfrac{1}{2}(x -\tan^{-1} x) + k= \bbox[#FFECB3,5px]{\dfrac{1}{2}\left((x^2+1)\tan^{-1} x - x\right) + k}$