Question

Вычислить неопределенные интегралы

Answer 1

$displaystyle int sqrt{1+2ln^2x}frac{lnxdx}{x}= int sqrt{1+2ln^2x}lnxd(lnx)=\= frac{1}{4}int sqrt{1+2ln^2x}d(1+2ln^2x)=frac{sqrt{(1+2ln^2x)^3}}{6}+C$

$displaystyle int frac{3x+4}{sqrt{(x+2)(x-1)}}dx=frac{3}{2}int frac{2x+1+frac{5}{3}}{sqrt{x^2+x-2}}dx=frac{3}{2}intfrac{d(x^2+x-2)}{sqrt{x^2+x-2}}+\+frac{5}{2}intfrac{d(x+frac{1}{2})}{sqrt{(x+frac{1}{2})^2-frac{9}{4}}}=3sqrt{x^2+x-2}+\+frac{5}{2}ln|(x+frac{1}{2})+sqrt{x^2+x-2}|+C\\\(x^2+x-2)'=2x+1$

$displaystyle int frac{lnxdx}{(2x-1)^3}=-frac{lnx}{4(2x-1)^2}+frac{1}{4}intfrac{dx}{x(2x-1)^2}=-frac{lnx}{4(2x-1)^2}+\+frac{1}{4}intfrac{dx}{x}-frac{1}{4}intfrac{d(2x-1)}{2x-1}+frac{1}{4}intfrac{d(2x-1)}{(2x-1)^2}=-frac{lnx}{4(2x-1)^2}+\+frac{1}{4}ln|frac{x}{2x-1}|-frac{1}{4(2x-1)}+C$
$u=lnx= textgreater du=frac{dx}{x}\dv=frac{dx}{(2x-1)^3}= textgreater v=-frac{1}{4(2x-1)^2}\\frac{1}{x(2x-1)^2}=frac{A}{x}+frac{B}{2x-1}+frac{C}{(2x-1)^2}=frac{1}{x}-frac{2}{2x-1}+frac{2}{(2x-1)^2}\1=A(4x^2-4x+1)+B(2x^2-x)+Cx\x^2|0=4A+2B= textgreater B=-2\x|0=-4A-B+C= textgreater C=2\x^0|1=A$