Question

Помогите с решением, даю 100 баллов

Answer 1

$10)\; \; \int \frac{dx}{\sqrt{4-9x^2}}=\int \frac{dx}{\sqrt{2^2-(3x)^2}}=arcsin\frac{3x}{2}+C\; ;\\\\11)\; \; \int \frac{x\, dx}{(x^2+1)^3}=\Big [\, t=x^2+1\; ,\; dt=2x\, dx\, \Big ]=\frac{1}{2}\int \frac{dt}{t^3}=\frac{1}{2}\int t^{-3}dt=\\\\=\frac{1}{2}\cdot \frac{t^{-2}}{-2}+C=\frac{1}{2}\cdot \frac{-1}{2t^2}+C=-\frac{1}{4(x^2+1)^2}+C\; ;$

$12)\; \int \frac{lnx}{x^3}\, dx=\Big [\, u=lnx,\; du=\frac{dx}{x},\; dv=\frac{dx}{x^3},\; v=\int \frac{dx}{x^3}=\frac{x^{-2}}{-2}=-\frac{1}{2x^2}\, \Big ]=\\\\=uv-\int v\, du=-\frac{lnx}{2x^2}+\frac{1}{2}\int \frac{dx}{x^3}=-\frac{lnx}{2x^2}-\frac{1}{2}\cdot (-\frac{1}{2x^2})+C=\\\\=-\frac{lnx}{2x^2}-\frac{1}{4x^2}+C\; .$

Answer 2

$\int\limits {\frac{1}{\sqrt{4-x^2} } } \,dx =arcsin(\frac{x}{2} )+C\\\int\limits {\frac{x}{(x^2+1)^3} } \,dx;x^2+1:=t=>\int\limits {\frac{1}{2t^3} } \,dt<=>\frac{1}{2}\int\limits {\frac{1}{2t^2} } \,dt=>\frac{1}{2} (-\frac{1}{2(x^2+1)^2} )=-\frac{1}{4(x^2+1)^2}+C\\\int\limits {\frac{ln(x)}{x^3} } \,dx<=>\int\limits {ln(x)}(\frac{1}{x^3})\,dx=>ln(x)(-\frac{1}{2x^2} )-\int\limits {-\frac{1}{2x^2} *\frac{1}{x} }\,dx<=>ln(x)(-\frac{1}{2x^2} )+\frac{1}{2} \int\limits {\frac{1}{x^3} } \,dx=-\frac{ln(x)}{2x^2} -\frac{1}{4x^2}+C$