Найти интегралы: Задание во вложении на фото

Решение в приложениях.

[latex]1)\; \; \int \frac{dx}{\sqrt{1-x^2}\cdot arcsin^4x} =\int \frac{dt}{t^4}=\int t^{-4}\cdot dt=\\\\=\frac{t^{-3}}{-3}+C=-\frac{1}{3arcsin^3x}+C\\\\2)\; \int (9-x^2)e^{3x}dx=[\, u=9-x^2,du=-2x\, dx,dv=e^{3x}dx,v=\frac{1}{3}e^{3x}]=\\\\=\frac{9-x^2}{3}e^{3x}+\frac{2}{3}\int xe^{3x}dx=[u=x,du=dx,dv=e^{3x}dx,v=\frac{1}{3}e^{3x}]=\\\\= \frac{9-x^2}{3}e^{3x}+\frac{2}{3}(\frac{x}{3}e^{3x}-\frac{1}{3}\int e^{3x}dx)= \frac{9-x^2}{3}e^{3x}+\frac{2}{9}xe^{3x}-\frac{2}{27}e^{3x}+C [/latex] [latex]3)\; \; \int \frac{2x^2-5x+1}{x^3-2x^2+x} dx=\int \frac{2x^2-5x+1}{x(x-1)^2} dx=I\\\\ \frac{2x^2-5x+1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2} \\\\2x^2-5x+1=A(x-1)^2+Bx(x-1)+Cx\\\\x=0,\; \; A=1\\\\x=1,\; \; C=-2\\\\x^2\, |\, 2=A+B\; \; \to \; \; B=2-A=2-1=1\\\\I=\int (\frac{1}{x}+\frac{1}{x-1}- \frac{2}{(x-1)^2} )dx=ln|x|+ln|x-1|+ \frac{2}{x-1} +C[/latex] [latex]4)\; \; \int \frac{dx}{8sin^2x-16sinxcosx} =\int \frac{dx/cos^2x}{8tg^2x-16tgx} =[t=tgx]=\int \frac{dt}{8(t^2-2t)} =\\\\=\frac{1}{8}\int \frac{dt}{(t-1)^2-1} =\frac{1}{8}\cdot \frac{1}{2}\cdot ln\left |\frac{(t-1)-1}{t-1+1}\right |+C=\frac{1}{16}\cdot ln\left |\frac{tgx-2}{tgx}\right |+C[/latex] [latex]5)\; \; \int \frac{\sqrt{x+1}-1}{(\sqrt[3]{x+1}+1)\sqrt{x+1}}dx=[x+1=t^6,x=t^6-1,dx=6t^5\, dt]=\\\\=\int \frac{(t^3-1)\cdot 6t^5}{(t^2+1)t^3} \cdot dt=6\cdot \int \frac{t^2(t^3-1)}{t^2+1} dt=6\cdot \int \frac{t^5-t^2}{t^2+1}dt=\\\\=6\cdot \int (t^3-t-1+\frac{t+1}{t^2+1})dt=6\frac{t^4}{4} -6\frac{t^2}{2}-6t+6\cdot \frac{1}{2}\int \frac{2t\, dt}{t^2+1}+6\int \frac{dt}{t^2+1}=\\\\=\frac{3}{2}\cdot \sqrt[6]{(x+1)^4}-3\sqrt[6]{(x+1)^2}-6\sqrt[6]{x+1}+3ln(t^2+1)+6arctgt+C[/latex] [latex]=\frac{3}{2}\sqrt[3]{(x+1)^2}-3\sqrt[3]{x+1}-6\sqrt[6]{x+1}+3ln(\sqrt[3]{x+1}+1)+\\\\+6arctg\sqrt[6]{x+1}+C\\\\6)\; \; \int \frac{\sqrt{(9-x^2)^3}}{x^4} dx=[x=3sint,\; 9-x^2=9cos^2t,\; dx=3cost\, dt]=\\\\=\int \frac{\sqrt{9^3cos^6t}\cdot 3cost}{3^4sin^4t} dt=\int \frac{3^4cos^4t}{3^4sin^4t} dt=\int ctg^4t\cdot dt=[ctgt=z,\; t=arcctgz,\\\\ dt=-\frac{dz}{1+z^2}\, ]=-\int \frac{z^4}{1+z^2}dz=-\int (z^2-1+\frac{1}{1+z^2})dz=\\\\=-\frac{z^3}{3}+z-arctgz+C=[/latex] [latex]=-\frac{1}{3}ctg^3(arcsin\frac{x}{3})+ctg(arcsin\frac{x}{3})-arctg(ctg(arcsin\frac{x}{3}))+C=\\\\=-\frac{1}{3}\cdot \frac{\sqrt{(9-x^2)^3}}{x^3}+\frac{\sqrt{9-x^2}}{x}-arctg\frac{\sqrt{9-x^2}}{x}+C[/latex]