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

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

$\displaystyle \int (3x+2)\sin x\,\mathrm{d}x$

$\displaystyle \int (3x+2)\sin x\,\mathrm{d}x \,\overset{\substack{ u=3x+2\quad u'=3\\ v'=\sin x\quad v=-\cos x\\ \phantom{\downarrow}}}{=}\, (3x+2)(-\cos x)-\int 3 (-\cos x)\,\mathrm{d}x= \bbox[#FFECB3,5px]{-(3x+2)\cos x + 3\sin x + k}$

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

$\displaystyle \int x^2\ln x\,\mathrm{d}x \,\overset{\substack{ u=\ln x\quad u'=\frac{1}{x}\\ v'=x^2\quad v=\frac{1}{3}x^3\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{3}x^3\ln x - \int\frac{1}{3}x^3\frac{1}{x}\,\mathrm{d}x= \dfrac{1}{3}x^3\ln x -\dfrac{1}{9}x^3 + k = \bbox[#FFECB3,5px]{\dfrac{1}{9}x^3(3\ln x - 1) + k}$

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

$\displaystyle \int \ln^3 x\,\mathrm{d}x \,\overset{\substack{ u=\ln^3 x\quad u'=3\ln^2 x\frac{1}{x}\\ v'=1\quad v=x\\ \phantom{\downarrow}}}{=}\, x\ln^3 x - \int 3\ln^2 x\frac{1}{x}x\,\mathrm{d}x \,\overset{\substack{ u=3\ln^2 x\quad u'=6\ln x\frac{1}{x}\\ v'=1\quad v=x\\ \phantom{\downarrow}}}{=}\,\\ \qquad\displaystyle x\ln^3 x - 3x\ln^2 x + \int 6\ln x\frac{1}{x}x\,\mathrm{d}x \,\overset{\substack{ u=6\ln x\quad u'=\frac{6}{x}\\ v'=1\quad v=x\\ \phantom{\downarrow}}}{=}\, x\ln^3 x - 3x\ln^2 x + 6x\ln x -\int 6\frac{1}{x}x\,\mathrm{d}x =\\ \qquad\displaystyle \bbox[#FFECB3,5px]{x(\ln^3 x - 3\ln^2 x + 6\ln x -6) + k}$

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

$\displaystyle \int x\cdot 3^{-x}\,\mathrm{d}x \,\overset{\substack{ u=x\quad u'=1\\ v'=3^{-x}\quad v=-\frac{1}{\ln 3}3^{-x}\\ \phantom{\downarrow}}}{=}\, -\frac{1}{\ln 3}x\cdot 3^{-x} - \int -\frac{1}{\ln 3}3^{-x}\,\mathrm{d}x= \bbox[#FFECB3,5px]{\left(-\frac{1}{\ln 3}x-\frac{1}{\ln^2 3}\right)3^{-x} + k}$