Question

Нужна помощь срочно!!! Нужно решить желательно с проверкой!!

1)int x^2dx/x^4+5x^2 +4

2)int sin^3 5xdx

3)int 5sqrtx -2x^3 +4/x^2 .dx

Answer 1

$1)\; \; \int \frac{x^2\, dx}{x^4+5x^2+4}=\int \frac{x^2\, dx}{(x^2-1)(x^2-4)}=\int \frac{x^2\, dx}{(x-1)(x+1)(x-2)(x+2)}=Q\\\\\frac{x^2}{(x-1)(x+1)(x-2)(x+2)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-2}+\frac{D}{x+2}\; ,\\\\x^2=A(x+1)(x-2)(x+2)+B(x-1)(x-2)(x+2)+\\\\+C(x-1)(x+1)(x+2)+D(x-1)(x+1)(x-2)\; ;\\\\x=-1:\; \; (-1)^2=B\cdot (-1-1)(-1-2)(-1+2)\; ,\; \; \; 1=B\cdot 6\; ;\\\\B=\frac{1}{6}\\\\x=1:\; \; 1^2=A(1+1)(1-2)(1+2)\; ,\; \; \; A=-\frac{1}{6}\\\\x=-2:\; \; (-2)^2=D(-2-1)(-2+1)(-2-2)\; ,\; \; D=-\frac{4}{12}=-\frac{1}{3}$

$x=2:\; \; 2^2=12\, C\; ,\; \; C=\frac{4}{12}=\frac{1}{3}\\\\Q=-\frac{1}{6}\int \frac{dx}{x-1}+\frac{1}{6}\int \frac{dx}{x+1}+\frac{1}{3}\int \frac{dx}{x-2}-\frac{1}{3}\int \frac{dx}{x+2}=\\\\=-\frac{1}{6}\, ln|x-1|+\frac{1}{6}\, ln|x+1|+\frac{1}{3}ln|x-2|-\frac{1}{3}\, ln|x+2|+C=\\\\=\frac{1}{6}\, ln\Big | \frac{x+1}{x-1}\Big |+\frac{1}{3}\, ln\Big | \frac{x-2}{x+2}\Big |+C\; ;$

$2)\; \; \int sin^35x\, dx=\int sin^25x\cdot sin5x\, dx=\int (1-cos^25x)\cdot sin5x\, dx=\\\\=\int sin5x\, dx-\int cos^25x\cdot \underbrace {sin5x\, dx}_{-1/5\cdot d(cos5x)}=-\frac{1}{5}cos5x+ \frac{1}{5}\cdot \frac{cos^35x}{3}+C\; ;\\\\\\3)\; \; \int \frac{5\sqrt{x}-2x^3+4}{x^2}\, dx=\int (5x^{-3/2}-2x+4x^{-2})dx=\\\\=\frac{5x^{-1/2}}{-1/2}-\frac{2x^2}{2}+\frac{4x^{-1}}{-1}+C=-\frac{10}{\sqrt{x}}-x^2-\frac{4}{x}+C\; .$