Определённые интегралы

Question 1

Question 2

Решите задачу:

$1)\; \; \int\limits^1_0\, \frac{x\, dx}{x^4+3}= \int\limits^1_0\,\frac{x\, dx}{(x^2)^2+3} =[\, t=x^2\; ,\; dt=2x\, dx\; ,\; t_1=0\; ,\; t_2=1]=\\\\= \frac{1}{2}\, \int\limits^1_0\, \frac{dt}{t^2+3}= \frac{1}{2\sqrt3}\, arctg \frac{t}{\sqrt3}\; \Big |_0^1=\frac{1}{2\sqrt3}\, arctg \frac{x^2}{\sqrt3}\; \Big |_0^1=\\\\= \frac{1}{2\sqrt3} \, (arctg \frac{1}{\sqrt3}-0)= \frac{1}{2\sqrt3}\cdot \frac{\pi }{6} = \frac{\pi }{12\sqrt3}$

$2)\; \; \int\limits^1_0 \, \frac{x^4dx}{(2-x^2)^{3/2}}=[\, x=\sqrt2sint\; ,dx=\sqrt2cost\, dt\; ,\; t=arcsin\frac{x}{\sqrt2}\; ,\\\\t_1=0\; ,\; t_2= \frac{\pi }{4}\; ]= \int\limits^{\frac{\pi}{4}}_0 \, \frac{4sin^4t\cdot \sqrt2cost\, dt}{\sqrt{(2-2sin^2t)^3}}= \int\limits^{ \frac{\pi}{4}}_0\, \frac{4\sqrt2\, sin^4t\, cost\, dt}{\sqrt{(2cos^2t)^3}} =\\\\= \int\limits^{\frac{\pi}{4}}_0\, \frac{4\sqrt2\, sin^4t\, cost\, dt}{2\sqrt2\, cos^3t} = 2\, \int\limits^{ \frac{\pi}{4}}_0\, \frac{(sin^2t)^2}{cos^2t}\, dt=2\, \int\limits^{\frac{\pi}{4}}_0\, \frac{(1-cos^2t)^2}{cos^2t}\, dt=$

$=2\, \int\limits^{\frac{\pi}{4}}_0\, \frac{1-2cos^2t+cos^4t}{cos^2t}\, dt=2 \int\limits^{\frac{\pi}{4}}_0( \frac{1}{cos^2t}-2+cos^2t)dt=\\\\=2(tgt-2t)\Big |_0^{\frac{\pi}{4}}+ \int\limits^{\frac{\pi}{4}}_0\, (1+cos2t)dt=\\\\=2(tg\frac{\pi}{4}-2\cdot \frac{\pi}{4})+(t+\frac{1}{2}\, sin2t)\Big |_0^{\frac{\pi }{4}}=2(1- \frac{\pi}{2})+\frac{\pi}{4}+\frac{1}{2}=\\\\=2-\pi + \frac{\pi }{4}+\frac{1}{2}=- \frac{3\pi }{4}+ \frac{5}{2}$

$3)\; \; \int\limits^1_0\, (3x-2)\, 3^{1-x}\, dx=[\, u=3x-2\; ,\; du=3\, dx\; ,\; dv=3^{1-x}dx, \\\\v=- \frac{3^{1-x}}{ln3}\; ]=(3x-2)\cdot \frac{3^{1-x}}{ln3}\; \Big |_0^1+\frac{3}{ln3}\, \int\limits^1_0\, 3^{1-x}\, dx=\\\\=\frac{1}{ln3}- \frac{2\cdot 3}{ln3}+ \frac{3}{ln3}\cdot \frac{-3^{1-x}}{ln3}\, \Big |_0^1=- \frac{5}{ln3}- \frac{3}{ln^23}\cdot (3^0-3)=-\frac{5}{ln3}+\frac{6}{ln^23}$