[latex] \int\limits ({x^3} - 3^x + \frac{ \sqrt{2} }{x} ) dx \int\limits ( \pi cos x - \frac{1}{cos^2 x} +10 ) dx \int\limits ( \frac{e^x sin^2 x + 3 sin^3 x + 1}{sin^2 x} ) dx \int\limits ( \sqrt[5]{x^2} - \frac{1}{ \sqrt{x^3...

[latex]\small \\ \int (x^3-3^x+{\sqrt2\over x}){\mathrm dx}={x^4\over4}-{3^x\over\ln3}+\sqrt{2}ln|x|+C\\\\ {\left ( {x^4\over4}-{3^x\over\ln3}+\sqrt{2}ln|x|+C \right )}'=x^3-3x+{\sqrt2\over x}\\\\ \int(\pi\cos{x}-{1\over\cos^2{x}}+10){\mathrm dx}=\pi\sin{x}-tg{x}+10x+C\\\\ {\left ( \pi\sin{x}-tg{x}+10x+C \right )}'= \pi\cos{x}-{1\over\cos^2x}+10\\\\ \int{e^x\sin^2{x}+3\sin^3{x}+1\over \sin^2{x}}{\mathrm dx}=\int e^x{\mathrm dx}+3\int \sin{x}{\mathrm dx}+\int{{\mathrm dx}\over \sin^2{x}}=e^x-3\cos{x}-ctg{x}+C\\\\ (e^x-3\cos{x}-ctg{x}+C)'=e^x+3\sin{x}+{1\over \sin^2{x}}[/latex][latex]\small \\ \int(\sqrt[5]{x^2}-{1\over \sqrt{x^3}}){\mathrm dx}={5\over7}\sqrt[5]{x^7}+{2\over \sqrt{x}}+C\\\\ \left ( {5\over7}\sqrt[5]{x^7}-{2\over \sqrt{x}}+C \right )'=x^{2\over5}-x^{-3\over2}=\sqrt[5]{x^2}-{1\over \sqrt{x^3}}[/latex]