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$\displaystyle \int x^2\cos(4x^3-2)\,\mathrm{d}x$

$\displaystyle \int x^2\cos(4x^3-2)\,\mathrm{d}x= \dfrac{1}{12}\int \cos(4x^3-2)12x^2\,\mathrm{d}x \,\overset{\substack{t=4x^3-2\\ \mathrm{d}t=12x^2\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{12}\int \cos t\,\mathrm{d}t= \dfrac{1}{12}\sin t + k \,\overset{\substack{t=4x^3-2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{12}\sin(4x^3-2) + k}$

$\displaystyle \int \dfrac{\mathrm{d}x}{a^2+x^2}$

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

$\displaystyle \int x(2x^2-2)^{11}\,\mathrm{d}x$

$\displaystyle \int x(2x^2-2)^{11}\,\mathrm{d}x= \dfrac{1}{4}\int (2x^2-2)^{11}4x\,\mathrm{d}x \,\overset{\substack{t=2x^2-2\\ \mathrm{d}t=4x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{4}\int t^{11}\,\mathrm{d}t= \dfrac{1}{4}\dfrac{1}{12}t^{12} + k \,\overset{\substack{t=2x^2-2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{48}(2x^2-2)^{12} + k}$

$\displaystyle \int x\sqrt[3]{4+2x^2}\,\mathrm{d}x$

$\displaystyle \int x\sqrt[3]{4+2x^2}\,\mathrm{d}x= \dfrac{1}{4}\int \sqrt[3]{4+2x^2}4x\,\mathrm{d}x \,\overset{\substack{t=4+2x^2\\ \mathrm{d}t=4x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \dfrac{1}{4}\int t^{\frac{1}{3}}\,\mathrm{d}t= \dfrac{1}{4}\dfrac{3}{4}t^{\frac{4}{3}} + k \,\overset{\substack{t=4+2x^2\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{3}{16}(4+2x^2)^{\frac{4}{3}} + k}$

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

$\displaystyle \int \sin^2 x\cos x\,\mathrm{d}x \,\overset{\substack{t=\sin x\\ \mathrm{d}t=\cos x\,\mathrm{d}x\\ \phantom{\downarrow}}}{=}\, \int t^2\,\mathrm{d}t= \dfrac{1}{3}t^3 + k \,\overset{\substack{t=\sin x\\ \phantom{\downarrow}}}{=}\, \bbox[#FFECB3,5px]{\dfrac{1}{3}\sin^3 x + k}$