Решите пожалуйста примеры. UPD: в 4 примере если не видно 2-x^2 а внизу 2x-1.

Question 1

Решите пожалуйста примеры.
UPD: в 4 примере если не видно 2-x^2 а внизу 2x-1.

Question 2

Решите задачу:

$U=z*tg(3x-2y)\\\frac{\delta U}{\delta x}=z*\frac{1}{cos^2(3x-2y)}*3=\frac{3z}{cos^2(3x-2y)}\\\frac{\delta U}{\delta y}=z*\frac{1}{cos^2(3x-2y)}*(-2)=\frac{-2z}{cos^2(3x-2y)}\\\frac{\delta U}{\delta z}=tg(3x-2y)$

$\int\limits^1_{-3}dx \int\limits^{2-x^2}_{2x-1}(x-y)dy= \int\limits^1_{-3}(x(2-x^2)-\frac{(2-x^2)^2}{2}-x(2x-1)+\\+\frac{(2x-1)^2}{2})dx=...\\\\\int\limits^{2-x^2}_{2x-1}(x-y)dy=xy-\frac{y^2}{2}|^{2-x^2}_{2x-1}=x(2-x^2)-\frac{(2-x^2)^2}{2}-x(2x-1)+\\+\frac{(2x-1)^2}{2}=x(2-x^2-2x+1)-\frac{1}{2}((2-x^2)^2-(2x-1)^2)=\\=x(-x^2-2x+3)-\frac{1}{2}((2-x^2-2x+1)(2-x^2+2x-1))=\\=(-x^2-2x+3)(x-\frac{1}{2}(-x^2+2x+1))=\\=(-x^2-2x+3)(x+\frac{x^2}{2}-x-\frac{1}{2})=(-x^2-2x+3)(\frac{x^2-1}{2})=$
$=\frac{1}{2}(-x^4-2x^3+4x^2+2x-3)\\...=\frac{1}{2}\int\limits^1_{-3}(-x^4-2x^3+4x^2+2x-3)dx=\\=\frac{1}{2}(-\frac{x^5}{5}-\frac{x^4}{2}+\frac{4x^3}{3}+x^2-3x)|^1_{-3}=\\\\=\frac{1}{2}(-\frac{1}{5}-\frac{1}{2}+\frac{4}{3}+1-3-\frac{243}{5}+\frac{81}{2}+36-9-9)=\\\\=\frac{1}{2}(-\frac{244}{5}+47+\frac{4}{3})=\frac{1}{2}(-48\frac{4}{5}+56+1\frac{1}{3})=\frac{1}{2}(9-\frac{7}{15})=\\\\=\frac{1}{2}(8\frac{8}{15})=\frac{1}{2}*\frac{128}{15}=\frac{64}{15}=4\frac{4}{15}$

$(1+x^2)y'+y=arctgx\\y=uv;y'=u'v+v'u\\(1+x^2)u'v+u((1+x^2)v'+v)=arctgx\\\begin{cases}(1+x^2)v'+v=0\\(1+x^2)u'v=arctgx\end{cases}\\(1+x^2)\frac{dv}{dx}+v=0|*-\frac{1}{dv}\\-\frac{1+x^2}{dx}=\frac{v}{dv}\\-\frac{dx}{1+x^2}=\frac{dv}{v}\\-\int\frac{dx}{1+x^2}=\int\frac{dv}{v}\\-arctgx=ln|v|\\v=e^{-arctgx}\\\frac{(1+x^2)e^{-arctgx}du}{dx}=arctgx|*\frac{dx}{(1+x^2)e^{-arctgx}}\\du=\frac{e^{arctgx}*arctgxdx}{1+x^2}\\\int du=\int\frac{e^{arctgx}*arctgxdx}{1+x^2}\\$
$\int e^{arctgx}*arctgx\ d(arctgx)=arctgx*e^{arctgx}}-\int e^{arctgx}d(arctgx)=\\=e^{arctgx}(arctgx-1)+C\\u=arctgx=\ \textgreater \ du=d(arctgx)\\dv=e^{arctgx}d(arctgx)=\ \textgreater \ v=e^{arctgx}\\\\u=e^{arctgx}(arctgx-1)+C\\y=(e^{arctgx}(arctgx-1)+C)e^{-arctgx}=arctgx-1+\frac{C}{e^{arctgx}}$