Сижу уже 6 час , помогитеее

А) $\\ \int\sqrt[4]{3-2sin{(3x)}}*cos{(3x)}\mathrm{d}x={1\over3}\int\sqrt[4]{3-2sin{(3x)}}*cos{(3x)}\mathrm{d}(3x)=\begin{vmatrix} u=3-2sin{(3x)}\\ du=-2cos{(3x)}d(3x) \end{vmatrix}= -{1\over6}\int\sqrt[4]{u}du=-{1\over6}*{4u^{5\over4}\over5}+C=-{2\over15}\sqrt[4]{(3-2sin{(3x)})^5}+C$ $\\$
б) $\\ \int{x\mathrm{d}x\over x^4+0,25}={1\over2}\int{\mathrm{d}x^2\over x^4+0,25}={1\over2}\int{\mathrm{d}x^2\over (x^2)^2+(0,5)^2}={1\over2}*{1\over 0,5}*arctg{x^2\over 0,5}+C=arctg{(2x^2)}+C$ $\\$
в) $\\ \int x*arcctg{x} \mathrm{d}x= \binom{u=arcctg{x},du=-{1\over1+x^2}dx}{dv=xdx,v={x^2\over2}}=\\\\\\ =v*u-\int v*du={x^2*arcctg{x}\over2}+{1\over2}\int{x^2\over1+x^2}\mathrm{d}x={x^2*arcctg{x}\over2}+{1\over2}\int{(1+x^2)-1\over1+x^2}\mathrm{d}x={x^2*arcctg{x}\over2}+{x\over2}-{1\over2}\int{1\over1+x^2}\mathrm{d}x={x^2*arcctg{x}\over2}+{x\over2}-{arctg{x}\over2}+C={1\over2}\begin{pmatrix} x^2*arcctg{x}+x-arctg{x} \end{pmatrix}+C$
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г) $\int_{-{1\over2}}^{0}{\mathrm{d}x\over \sqrt{1-x^2}}=\arcsin{x}|_{-{1\over2}}^{0}=\arcsin(0)-(\arcsin(-{1\over2}))=0+\arcsin{{1\over2}}={\pi\over6}$