Вычислить интегралы, пользуясь формулой Ньютона-Лейбница.

Question 1

Question 2

Решите задачу:

$\int _0^1\frac{6x+5}{\sqrt{x^2-4x+8}}dx=\int _0^1\frac{6x+5}{\sqrt{(x-2)^2+4}}dx=\\\\=[\, t=x-2,x=t+2,dx=dt,t_1=0-2=-2,t_2=1-2=-1\, ]=\\\\=\int _{-2}^{-1}\frac{6(t+2)+5}{\sqrt{t^2+4}}dt=\int _{-2}^{-1}\frac{6t+17}{\sqrt{t^2+4}}dt=3\cdot \int_{-2}^{-1}\frac{2t\cdot dt}{\sqrt{t^2+4}}+17\cdot \int _{-2}^{-1}\frac{dt}{\sqrt{t^2+4}}=\\\\=3\cdot 2\sqrt{t^2+4}|_{-2}^{-1}+17\cdot ln|t+\sqrt{t^2+4}||_{-2}^{-1}=\\\\=6(\sqrt5-\sqrt8)+17(ln(\sqrt3-1)-ln(\sqrt8-2)).$

$2)\; \int _0^{\frac{\pi}{2}}(x-2)sin7x\, dx=[\, u=x-2,\; du=dx;\\\\ dv=sin7x\, dx,\; v=\int dv=-\frac{1}{7}cos7x\; ; \; \; \int u\cdot dv=uv-\int v\cdot du\, ]=\\\\=-\frac{1}{7}(x-2)cos7x|_0^\frac{\pi}{2}}-\int _0^{\frac{\pi}{2}}(-\frac{1}{7})cos7x\, dx=\\\\=-\frac{1}{7}\cdot ((\frac{\pi}{2}-2)cos\frac{7\pi}{2}+2cos0)+\frac{1}{7}\cdot \frac{1}{7}sin7x|_0^{\frac{\pi}{2}}=[\, cos\frac{7\pi}{2}=0,cos0=1]\\\\=-\frac{1}{7}\cdot 2+\frac{1}{49}(sin\frac{7\pi}{2}-sin0)=[\, sin\frac{7\pi}{2}=-1,\; sin=0\, ]=$

$=-\frac{2}{7}-\frac{1}{49}=-\frac{15}{49}\\\\3)\; \int_1^2\frac{3x+5}{x(x^2-4x+8)}dx=I\\\\\frac{3x+5}{x(x^2-4x+8)}=\frac{A}{x}+\frac{Bx+C}{x^2-4x+8}=\frac{A(x^2-4x+8)+(Bx+C)\cdot x}{x(x^2-4x+8)}\\\\3x+5=Ax^2-4Ax+8A+Bx^2+Cx\\\\x^2\; \; |A+B=0\; ,\; \; B=-A=-\frac{5}{8};\\\\x\; \; \; |-4A+C=3\; ,\; \; C=3+4\cdot \frac{5}{8}=\frac{11}{2};\\\\x^0\; \; |8A=5\; ,\; \; \; A=\frac{5}{8}.$

$I=\frac{5}{8}\int _1^2\frac{dx}{x}+\int_1^2\frac{-\frac{5}{8}x+\frac{11}{2}}{x^2-4x+8}\cdot \frac{-8}{-8}\cdot dx=ln|x||_1^2-\frac{1}{8}\int _1^2\frac{5x-44}{(x-2)^2+4}dx=\\\\=[\, t=x-2,x=t+2,dx=dt,t_1=-1,t_2=0\, ]=\\\\=ln2-ln1-\frac{1}{8}\int _{-1}^0\frac{5t-34}{t^2+4}dt=\\\\=ln2-\frac{1}{8}\cdot (\frac{5}{2}\int _{-1}^0\frac{d(t^2+4)}{t^2+4}-34\cdot 4\cdot \int _{-1}^0\frac{dt}{t^2+4} )=\\\\=ln2-\frac{5}{16}ln|t^2+4||_{-1}^0-17\cdot \frac{1}{2}arctg\frac{t}{2}|_{-1}^0=\\\\=ln2-\frac{5}{16}(ln4-ln5)-$

$-\frac{17}{2}(arctg0-arctg(-\frac{1}{2}))=ln2-\frac{5}{16}\cdot ln\frac{4}{5}-\frac{17}{2}arctg\frac{1}{2}.$