Question

Решить неопределенный интеграл, заранее большое спасибо

Answer 1

$int frac{dx}{cos^2x(3tgx+1)}=frac{1}{3}intfrac{d(3tgx+1)}{(3tgx+1)}=frac{1}{3}ln|3tgx+1|+C\\(frac{1}{3}ln|3tgx+1|+C)'=frac{1}{3}*frac{1}{3tgx+1}*frac{3}{cos^2x}=frac{1}{cos^2x(3tgx+1)}$


$int xarcsinxdx=frac{x^2}{2}arcsinx-frac{1}{2}intfrac{x^2dx}{sqrt{1-x^2}}=\=frac{x^2}{2}arcsinx+frac{x}{4}sqrt{1-x^2}-frac{1}{4}arcsinx+C=\=(frac{x^2}{2}-frac{1}{4})arcsinx+frac{x}{4}sqrt{1-x^2}+C\u=arcsinx= textgreater du=frac{dx}{sqrt{1-x^2}}\dv=xdx= textgreater v=frac{x^2}{2}$
$intfrac{x^2dx}{sqrt{1-x^2}}=-intfrac{(1-x^2-1)dx}{sqrt{1-x^2}}=-intsqrt{1-x^2}dx+intfrac{dx}{sqrt{1-x^2}}=\-xsqrt{1-x^2}-intfrac{x^2dx}{sqrt{1-x^2}}+arcsinx\2intfrac{x^2dx}{sqrt{1-x^2}}=-xsqrt{1-x^2}+arcsinx\intfrac{x^2dx}{sqrt{1-x^2}}=-frac{x}{2}sqrt{1-x^2}+frac{1}{2}arcsinx$
$intsqrt{1-x^2}dx=xsqrt{1-x^2}+intfrac{x^2dx}{sqrt{1-x^2}}\u=sqrt{1-x^2}= textgreater du=-frac{xdx}{sqrt{1-x^2}}\dv=dx= textgreater v=x$
$((frac{x^2}{2}-frac{1}{4})arcsinx+frac{x}{4}sqrt{1-x^2}+C)'=\=xarcsinx+frac{frac{x^2}{2}-frac{1}{4}}{sqrt{1-x^2}}+frac{1}{4}(sqrt{1-x^2}-frac{2x^2}{2sqrt{1-x^2}})=\=xarcsinx+frac{2x^2-1}{4sqrt{1-x^2}}+frac{1}{4}*frac{2-2x^2-2x^2}{2sqrt{1-x^2}}=xarcsinx+(frac{2x^2-1+1-2x^2}{4sqrt{1-x^2}})=\=xarcsinx$


$intfrac{x^3dx}{x^4+2}=frac{1}{4}intfrac{d(x^4+2)}{x^4+2}=frac{1}{4}ln|x^4+2|+C\\(frac{1}{4}ln|x^4+2|+C)'=frac{1}{4}*frac{4x^3}{x^4+2}=frac{x^3}{x^4+2}$