Интегралы, помогите сколько сможете, желательно все.

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

$2); int frac{x, dx}{sqrt{1-2x^4}}=frac{1}{2}int frac{2x, dx}{sqrt{1-2(x^2)^2}}=[, t=x^2,; dt=2x, dx, ]==frac{1}{2}int frac{dt}{sqrt{1-2t^2}}=frac{1}{2}int frac{dt}{sqrt2cdot sqrt{frac{1}{2}-t^2}}=frac{1}{2sqrt2}cdot arcsinfrac{t}{frac{1}{sqrt2}}}+C==frac{1}{2sqrt2}arcsin(sqrt2x^2)+C$

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

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

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

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

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