Question

помогите пожалуйста
нужно подробное решение

Answer 1

$\int\limits^1_0 \frac{8x+2}{\sqrt{x^2+6x+10}}= 4\int\limits^1_0 \frac{2x+6-5\frac{1}{2}}{\sqrt{x^2+6x+10}}=4\int\limits^1_0\frac{d(x^2+6x+10)}{\sqrt{x^2+6x+10}}-22\int\limits^1_0\frac{d(x+3)}{\sqrt{(x+3)^2+1}}=\\=8\sqrt{x^2+6x+10}|^1_0-22ln|x+3+\sqrt{{(x+3)^2+1}}||^1_0\\8\sqrt{17}-8\sqrt{10}-22ln|4+\sqrt{17}|+22ln|3+\sqrt{10}|=\\8(\sqrt{17}-\sqrt{10})+22ln|\frac{3+\sqrt{10}}{4+\sqrt{17}}|\approx1,6\\\\(x^2+6x+10)'=2x+6$

$\int\limits^\pi_\frac{\pi}{2}xsin(3x-1)dx=-\frac{x}{3}cos(3x-1)|^{\pi}_{\frac{\pi}{2}}+\frac{1}{3}\int\limits^\pi_\frac{\pi}{2}cos(3x-1)dx=\\=-\frac{x}{3}cos(3x-1)|^{\pi}_{\frac{\pi}{2}}+\frac{1}{9}sin(3x-1)|^{\pi}_{\frac{\pi}{2}}=-\frac{\pi}{3}cos(3\pi-1)+\\+\frac{\pi}{6}cos(\frac{3\pi}{2}-1)+\frac{1}{9}sin(3\pi-1)-\frac{1}{9}sin(\frac{3\pi}{2}-1)\approx\\\approx0,28\\u=x=\ \textgreater \ du=dx\\dv=sin(3x-1)dx=\ \textgreater \ v=-\frac{1}{3}cos(3x-1)$

$\int\limits^1_0 \frac{3-2x}{(x+1)(x^2+6x+10)}dx= \int\limits^1_0 \frac{dx}{x+1}-\frac{1}{2}\int\limits^1_0\frac{d(x^2+6x+10)}{x^2+6x+10}-4\int\limits^1_0\frac{1}{(x+3)^2+1}=\\=ln|x+1||^1_0-\frac{1}{2}ln|x^2+6x+10||^1_0-4arctg(x+3)|^1_0=\\=ln|2|-\frac{1}{2}ln|\frac{17}{10}|-4arctg(4)+4arctg(3)\approx\\\approx0,69-0,27-5,30+5\approx0,12$
$\frac{3-2x}{(x+1)(x^2+6x+10)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+6x+10}=\frac{1}{x+1}-\frac{x+7}{x^2+6x+10}\\3-2x=A(x^2+6x+10)+B(x^2+x)+C(x+1)\\x^2|0=A+B=\ \textgreater \ B=-A\\x^1|-2=6A+B+C=\ \textgreater \ -2=5A+C\\x^0|3=10A+C\\-5=-5A\\A=1\\B=-1\\C=-7$