Решить интегралы, желательно подробно.

[latex]11)\;\int(x^2+3)^5xdx=\left(\begin{array}{c}u=x^2+3\\du=2xdx\end{array}\right)=\frac12\int u^5du=\frac12\cdot\frac{u^6}6=\frac{u^6}{12}=\\=\frac{(x^2+3)^6}{12}\\12)\;\int\frac{(x^2-7)^3x}2dx=\frac12\int(x^2-7)^3xdx=\left(\begin{array}{c}u=x^2-7\\du=2xdx\end{array}\right)=\\=\frac14\int u^3dx=\frac14\cdot\frac{u^4}4=\frac{u^5}{16}=\frac{(x^2-7)^5}{16}\\13)\;\int\sqrt{(x^4-3)^3}x^3dx=\left(\begin{array}{c}u=x^4-1\\du=4x^3dx\end{array}\right)=\frac14\int u^{\frac32}du=\frac14\cdot\frac25u^{\frac52}=[/latex] [latex]=\frac{u^\frac52}{20}=\frac{\sqrt{(x^4-1)^5}}{20}[/latex] [latex]14)\;\int\frac{\cos xdx}{5\sin x+2}=\left(\begin{array}{c}u=5\sin x+2\\du=5\cos xdx\end{array}\right)=\frac15\int\frac{du}u=\frac15\ln u=\\=\frac15\ln(5\sin x+2)\\15)\;\int\frac{dx}{\sqrt{(3x-2)^3}}=\left(\begin{array}{c}u=3x-2\\du=3dx\end{array}\right)=\frac13\int\frac{du}{u^{\frac32}}=\frac13\cdot\frac{-2}{u^{\frac13}}=-\frac2{3u^{\frac12}}=\\=-\frac2{3\sqrt{3x-2}}\\16)\;\int\sqrt{e^x+1}e^xdx=\left(\begin{array}{c}u=e^x+1\\du=e^xdx\end{array}\right)=\int\sqrt udu=\frac23 u^{\frac32}=[/latex] [latex]=\frac{2\sqrt{(e^x+1)}}{3}\\17)\;\int\sqrt[3]{(5-2x)^2}dx=\left(\begin{array}{c}u=5-2x\\du=-2dx\end{array}\right)=-\frac12\int u^{\frac23}du=-\frac12\cdot\frac35 u^{\frac53}=\\=-\frac{3u^{\frac53}}{10}=-\frac{3\sqrt[3]{(5-2x)^5}}{10}[/latex]