Помогите решить мнтегралы

Question 1

Question 2

Решите задачу:

$8)\; \; \int \frac{\sqrt{(1+arctg2x)^3}}{4x^2+1}dx=[\, u=1+arctg2x,\; du=\frac{2\,dx}{1+4x^2}\, ]=\\\\=\frac{1}{2}\int \sqrt{u^3}\, du=\frac{1}{2}\cdot \frac{u^{5/2}}{5/2}+C=\frac{\sqrt{(1+arctg2x)^5}}{5}+C\\\\9)\; \; \int 5^{tgx}\cdot \frac{dx}{cos^2x}=[u=tgx]=\int 5^{u}\cdot du=\frac{5^{u}}{ln5}+C=\frac{5^{tgx}}{ln5}+C\\\\10)\; \; \int \frac{cos(ln3x)}{x}dx=[\, u=ln3x,\; du=\frac{1}{3x}\cdot 3dx=\frac{dx}{x}\, ]=\\\\=\int cosu\cdot du=sinu+C=sin(ln3x)+C\\\\12)\; \; \int ctg(e^{2x}+5)\cdot e^{2x}dx=[\, u=e^{2x}+5,\; du=2e^{2x}dx\, ]=$

$=\frac{1}{2}\int ctgu\cdot du=\frac{1}{2}\int \frac{cosu\, du}{sinu}=\frac{1}{2}\int \frac{d(sinu)}{sinu}=\frac{1}{2}\cdot ln|sinu|+C=\\\\=\frac{1}{2}\cdot ln|sin(e^{2x}+5)|+C\\\\13)\; \; \int \frac{x^2\, dx}{\sqrt{9+x^6}}=\int \frac{x^2\, dx}{\sqrt{9+(x^3)^2}}=[\, u=x^3,\; du=3x^2dx\, ]=\\\\=\frac{1}{3}\int \frac{du}{\sqrt{9+u^2}}=\frac{1}{3}\cdot ln|u+\sqrt{9+u^2}|+C=\frac{1}{3}\cdot ln|x^3+\sqrt{9+x^6}|+C\\\\7)\; \; \int \frac{dx}{sin(4x-1)}=[u=4x-1,\; du=4dx\, ]=\frac{1}{4}\int \frac{du}{sinu}=$

$=[\, t=tg\frac{u}{2},\; sinu=\frac{2t}{1+t^2},\; du=\frac{2\, dt}{1+t^2} \, ]=\frac{1}{4}\int \frac{\frac{2\, dt}{1+t^2}}{\frac{2t}{1+t^2}}=\frac{1}{4}\int \frac{dt}{t}=\\\\=\frac{1}{4}\cdot ln|t|+C=\frac{1}{4}\cdot ln|tg\frac{u}{2}|+C=\frac{1}{4}\cdot ln|tg\frac{4x-1}{2}|+C\\\\11)\; \; \int \frac{dx}{\sqrt{x}\sqrt{5-x}}=\int \frac{dx}{\sqrt{5x-x^2}}=\\\\=[5x-x^2=-(x^2-5x)=-((x-\frac{5}{2})^2-\frac{25}{4})=\frac{25}{4}-(x-\frac{5}{2})^2,\, \\\\t=x-\frac{5}{2},\; dt=dx\, ]=\int \frac{dt}{\sqrt{\frac{25}{4}-t^2}}=$

$=arcsin \frac{t}{5/2}+C=arcsin\frac{2(x-\frac{5}{2})}{5} +C=arcsin\frac{2x-5}{5}+C$