Помогите пожалуйста, очень надо 3,4,6,13,16

Question 1

Question 2

Решите задачу:

$3)\; \; \int \frac{dx}{(7x-4)^2}=\int (7x-4)^{-2}dx=\\\\=[\, t=7x-4\; ,\; dt=(7x-4)'dx=7dx\; ,\; dx=\frac{dt}{7}\, ]=\\\\=\frac{1}{7}\int t^{-2}dt=\frac{1}{7}\cdot \frac{t^{-1}}{-1}+C=-\frac{1}{7t}+C=-\frac{1}{7(7x-4)}+C\\\\4)\; \; \int \frac{2x-7}{x^2-7x+5}dx=\\\\=[\, t=x^2-7x+5\; ,\; dt=(x^2-7x+5)'dx=(2x-7)dx]=\\\\=\int \frac{dt}{t}=ln|t|+C=ln|x^2-7x+5|+C\\\\6)\; \; \int \frac{dx}{x\sqrt{lnx}}=\int \frac{\frac{dx}{x}}{\sqrt{lnx}}=[\, t=lnx\; ,\; dt=(lnx)'dx=\frac{dx}{x}\, ]=\\\\=\int \frac{dt}{\sqrt{t}}=2\sqrt{t}+C=2\sqrt{lnx}+C$

$13)\; \; \int \frac{3x-4}{x^2-4x+13}dx=\int \frac{3x-4}{(x-2)^2+9}=[\, t=x-2,\; x=t+2,\; dt=dx\, ]=\\\\=\int \frac{3(t+2)-4}{t^2+9}dt=\int \frac{3t+2}{t^2+9}dt=3\int \frac{t\, dt}{t^2+9}+2\int \frac{dt}{t^2+3^2}=\\\\=\frac{3}{2}\underbrace {\int \frac{2t\, dt}{t^2+9}}_{u=t^2+9,du=2tdt}+2\cdot \frac{1}{3}\cdot arctg\frac{t}{3}=\frac{3}{2}\int \frac{du}{u}+\frac{2}{3}arctg\frac{x-2}{3}=\\\\= \frac{3}{2}\cdot ln|u|+\frac{2}{3}arctg\frac{x-2}{3}+C=\frac{3}{2}\cdot ln|x^2-4x+13|+\frac{2}{3}arctg\frac{x-2}{3}+C$

$16)\; \; \int \frac{3x+2}{\sqrt{5+12x-9x^2}}dx=[-(9x^2-12x-5)=-((3x-2)^2-9)\, ]=\\\\=\int \frac{3x+2}{\sqrt{9-(3x-2)^2}}dx=[t=3x-2,\; dt=3dx,\; 3x=t+2\, ]=\\\\=\int \frac{t+4}{\sqrt{9-t^2}}dt=-\frac{1}{2} \int \frac{-2t\, dt}{\sqrt{9-t^2}}+4\int \frac{dt}{\sqrt{3^2-t^2}}=[u=9-t^2,\; dt=-2t\, dt]=\\\\=-\frac{1}{2}\int \frac{du}{\sqrt{u}}+4\cdot arcsin\frac{t}{3}=-\frac{1}{2}\cdot 2\sqrt{u}+4arcsin \frac{3x-2}{3}+C=\\\\=-\sqrt{5+12x-9x^2}+4arcsin \frac{3x-2}{2}+C$