Question

Интегралы, помогите сколько сможете, желательно все.

Answer 1

$1)\; \int \frac{dx}{x\sqrt{1+ln^2x}}=[\, t=lnx,\; dt=\frac{dx}{x}\, ]=\int \frac{dt}{\sqrt{1+t^2}}=\\\\=ln|t+\sqrt{1+t^2}|+C=ln|lnx+\sqrt{1+lnx}|+C$

$2)\; \int \frac{x\, dx}{\sqrt{1-2x^4}}=\frac{1}{2}\int \frac{2x\, dx}{\sqrt{1-2(x^2)^2}}=[\, t=x^2,\; dt=2x\, dx\, ]=\\\\=\frac{1}{2}\int \frac{dt}{\sqrt{1-2t^2}}=\frac{1}{2}\int \frac{dt}{\sqrt2\cdot \sqrt{\frac{1}{2}-t^2}}=\frac{1}{2\sqrt2}\cdot arcsin\frac{t}{\frac{1}{\sqrt2}}}+C=\\\\=\frac{1}{2\sqrt2}arcsin(\sqrt2x^2)+C$

$3)\; \int \frac{x}{2^{x}}dx=\int x\cdot 2^{-x}\, dx=\\\\=[\, u=x,\; du=dx,\; dv=2^{-x}dx,\; v=-\frac{2^{-x}}{ln2}\, ]=\\\\=-\frac{x2^{-x}}{ln2}+\frac{1}{ln2}\int 2^{-x}\, dx=-\frac{x2^{-x}}{ln2}-\frac{2^{-x}}{ln^22}+C$

$4)\; \int \frac{4x-1}{x^2-6x+11}dx=\int \frac{4x-1}{(x-3)^2+2}dx=\\\\=[t=x-3,\; x=t+3,\; dx=dt\, ]=\\\\=\int \frac{4t+11}{t^2+2}=2\int \frac{2t\, dt}{t^2+2}+11\int \frac{dt}{t^2+2}=2\int \frac{d(t^2+2)}{t^2+2}+11\cdot \frac{1}{\sqrt2}arctg\frac{t}{\sqrt2}+C=\\\\=2ln|t^2+2|+\frac{11}{\sqrt2}arctg\frac{x-3}{\sqrt2}+C=\\\\=2ln|x^2-6x+11|+\frac{11}{\sqrt2}arctg\frac{x-3}{\sqrt2}+C$

$5)\; \int \frac{3x^2-x-3}{x^3-x}dx=\int \frac{3x^2-x-3}{x(x-1)(x+1)}dx=Q\\\\\frac{3x^2-x-3}{x(x-1)(x+1)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x+1}\\\\x=0\; \to \; A=\frac{-3}{-1}=3\\\\x=1\; \to \; B=\frac{3-1-3}{1\cdot 2}=-\frac{1}{2}\\\\x=-1\; \to \; C=\frac{3+1-3}{-1\cdot (-2)}=\frac{1}{2}\\\\Q=3\int \frac{dx}{x}-\frac{1}{2}\int \frac{dx}{x-1}+\frac{1}{2}\int \frac{dx}{x+1}=3ln|x|-\frac{1}{2}ln|x-1|+\frac{1}{2}ln|x+1|+C$

$7)\; \int \frac{dx}{2sin^2x-3cos^2x}=\int \frac{dx/cos^2x}{2tg^2x-3}=\int \frac{d(tgx)}{2tg^2x-3}=\frac{1}{2}\int \frac{d(tgx)}{tg^2x-\frac{3}{2}}=\\\\=[\int \frac{dt}{t^2-\frac{3}{2}}=\frac{\sqrt2}{2\sqrt3}ln|\frac{t-\frac{\sqrt3}{\sqrt2}}{t+\frac{\sqrt3}{\sqrt2}}|+C]=\frac{1}{2}\cdot \frac{1}{\sqrt6}ln|\frac{\sqrt2tgx-\sqrt3}{\sqrt2tgx+\sqrt3}|+C$

$6)\; \int ctg^33x\, dx=\\\\=[t=ctg3x,\; 3x=arcctgt,\; x=\frac{1}{3}arcctgt,\; dx=\frac{1}{3}\cdot \frac{-dt}{t^2+1}]=\\\\=\int \frac{-t^3\, dt}{3(t^2+1)}=-\frac{1}{3}\int (t-\frac{t}{t^2+1}=-\frac{1}{6}t^2+\frac{1}{6}\int \frac{d(t^2+1)}{t^2+1}=\\\\=-\frac{1}{6}ctg^23x+\frac{1}{6}ln|ctg^23x+1|+C$