Verified answer

Объяснение:

a)

[tex]\int\limits {\frac{(x+1)^3}{x^2} } \, dx =\int\limits {\frac{x^3+3x^2+3x+1}{x^2} } \, dx =\int\limits {(x+3+\frac{3}{x} +\frac{1}{x^2}) } \, dx =\\=\int\limits {x} \, dx+\int\limits {3} \, dx +3\int\limits {\frac{dx}{x} } \,+\int\limits {x^{-2}} \, dx= \frac{x^2}{2} +3x+3*ln(|x|)-\frac{1}{x}+C=\\ =3*ln(|x|)+\frac{x^3+6x^2-2}{2x}+C.[/tex]

б)

[tex]\int\limits {(\frac{1}{\sqrt[3]{x^2}} -4x^{-3}} +cos(3x))} \, dx=\int\limits {x^{-\frac{2}{3} } \, dx -\int\limits {4x^{-3}} \, dx+\int\limits cos(3x) {x} \, dx =\\[/tex]

[tex]=3\sqrt[3]{x} -4*(-\frac{1}{2x^2} )+\frac{sin(3x)}{3}+C=3\sqrt[3]{x} +\frac{2}{x^2}+\frac{sin(3x)}{3}+C.[/tex]

в)

[tex]\int\limits {\frac{dx}{x^2+6x+10} } \, =\int\limits {\frac{dx}{x^2+2*x*3+3^2+1} } \, dx =\int\limits {\frac{dx}{(x+3)^2+1} } \,=\left | {{u=x+3} \atop {du=dx}} \right. \left | =\int\limits {\frac{du}{u^2+1} } \, dx =\\[/tex]

[tex]=arctg(u)+C=arctg(x+3)+C.[/tex]