Помогите пожалуйста. Интегралы.

Question 1

Question 2

Решите задачу:

$1)\; \; \int \frac{5x-16}{\sqrt{x^2+2x+17}}dx=\int \frac{5x-16}{\sqrt{(x+1)^2+16}}dx=\\\\=[\; t=x+1,\; x=t-1dt=dx\; ]=\int \frac{5t-21}{\sqrt{t^2-16}}dt=\\\\=\frac{5}{2}\int \frac{2t\, dt}{\sqrt{t^2-16}} -21\int \frac{dt}{\sqrt{t^2-16}}=\\\\=[\; \int \frac{2t\, dt}{\sqrt{t^2-16}}=\int \frac{d(t^2-16)}{\sqrt{t^2-16}}=\int \frac{dz}{\sqrt{z}}=2\sqrt{z}+C=2\sqrt{t^2-16}+C\; ]=\\\\=\frac{5}{2}\cdot 2\sqrt{t^2-16}-21\cdot arcsin\frac{t}{4}+C=\\\\=5\cdot \sqrt{x^2+2x+17}-21\cdot arcsin\frac{x+1}{4}+C$

$2)\; \; \int \frac{2x-1}{\sqrt{x^2-4x+5}}dx=\int \frac{2x-1}{\sqrt{(x-2)^2+1}}dx=\\\\=[\; t=x-2\; ,\; x=t+2,\; dt=dx\; ]=\int \frac{2t+3}{\sqrt{t^2+1}} dt=\\\\=\int \frac{2t\; dt}{\sqrt{t^2+1}} +3\int \frac{dt}{\sqrt{t^2+1}}=2\sqrt{t^2+1}+3\cdot ln|t+\sqrt{t^2+1}|+C=\\\\=2\sqrt{x^2-4x+5}+3\cdot ln|x-2+\sqrt{x^2-4x+5}|+C$

$3)\; \; \int \frac{17x+3}{x^2+8x+32} dx=\int \frac{17x+3}{(x+4)^2+16}dx=[\; t=x+4,dt=dx,\\\\x=t-4\; ]=\int \frac{17t-65}{t^2+16}dt= \frac{17}{2} \int \frac{2t\; dt}{t^2+16}-65\cdot \int \frac{dt}{t^2+16} =\\\\=[\; \int \frac{2t\; dt}{t^2+16}=\int \frac{d(t^2+16)}{t^2+16}=\int \frac{dz}{z}=ln|z|+C=ln|t^2+16|+C\; ]=\\\\=\frac{17}{2}\cdot ln|t^2+16|-65\cdot \frac{1}{4}\cdot arctg\frac{t}{4}+C=\\\\\frac{17}{2}\cdot ln(x^2+8x+32)-\frac{65}{4}\cdot arctg\frac{x+4}{4}+C$

$4)\; \; \int \frac{x+10}{-x^2+8x+4}dx=-\int \frac{x+10}{x^2-8x-4}dx=-\int \frac{x+10}{(x-4)^2-20} dx=\\\\=[\; t=x-4\; dt=dx\; ,\; x=t+4\; ]=-\int \frac{t+14}{t^2-20} \, dt=\\\\=-\frac{1}{2}\int \frac{2t\; dt}{t^2-20} -14\cdot \int \frac{dt}{t^2-20} =-\frac{1}{2}\cdot ln|t^2-20|-\\\\-14\cdot \frac{1}{2\sqrt{20}}\cdot ln\Big | \frac{t-\sqrt{20}}{t+\sqrt{20}} \Big |+C=\\\\=-\frac{1}{2}\cdot ln|x^2-8x-4|-\frac{7}{2\sqrt5}\cdot ln\Big | \frac{x-4-2\sqrt5}{x-4+2\sqrt5} \Big |+C$