Question

Вычислить неопределенные интегралы

Answer 1

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

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

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