Решить интеграл
$1) intlimits frac{dx}{5+9x} 2) intlimits ( sqrt[4]{x}- frac{2}{ sqrt[3]{x} }+ x^{4} )dx 3)intlimits xsin( x^{2} +1)dx 4)intlimits(3x-8)sinxdx 5)intlimits frac{5x+1}{ x^{2} +7}dx 6)intlimits x^{2} sinx$

1) $frac{1}{9} intlimits { frac{d(9x+5)}{9x+5} }= frac{1}{9}ln|9x+5|+C$
2) $intlimits ({x^{1/4}-2x^{-1/3}+x^4}) , dx = frac{x^{(5/4)}}{5/4}- frac{2x^{(2/3)}}{2/3}+ frac{x^5}{5}+C$
3) $frac{1}{2} intlimits {sin(x^2+1)} , d(x^2+1)= -cos( x^{2} +1)+C$
5) $intlimits { frac{5x}{ x^{2} +7} } , dx + intlimits { frac{dx}{ x^{2} +7} } = frac{5}{2}ln( x^{2} +7)+ frac{1}{ sqrt{7} }arctg frac{x}{ sqrt{7} }+C$
6)по частям: u=x² du=2xdx
dv=sinx v=-cosx
$- x^{2} cosx+2 intlimits {xcosx} , dx =$
u=x du=dx
dv=cosx v=sinx
= $- x^{2} cosx+xsinx- intlimits {sinx} , dx =-x^2cosx+xsinx+cosx+C$
4) $intlimits {3xsinx} , dx - intlimits {8sinx} , dx =-3xcosx+3 intlimits {cosx} , dx +8cosx=$
= $-3xcosx+3sinx+8cosx+C$