高数求助

(1)∫xcos^2xdx

=∫x·(1+cos2x)/2 dx

=1/2·(∫xdx + ∫xcos2x dx)

=1/4·[x^2 + ∫xd(sin2x)]

=1/4·x^2 + 1/4·x·sin2x - 1/4·∫d(sin2x) dx

=1/4·x^2 + 1/4·x·sin2x + 1/8·cos2x + C

(2)∫ln(1+x^2)dx

=x·ln(1+x^2) -∫xd[ln(1+x^2)]

=x·ln(1+x^2) -∫2x^2/(1+x^2) dx

=x·ln(1+x^2) - 2x + 2arctanx + C

(3)∫(2^x+3^x)^2dx

=∫(4^x + 2·6^x + 9^x)dx

=4^x/ln4 + 2·6^x/ln6 + 9^x/ln9 + C

(注：应该是2次方吧，你再核对一下题目；如果是3次方，处理方法类似，仍然展开再代入积分公式)

(4)∫xsinxcosxdx

=1/2·∫xsin2xdx

= -1/4·∫xd(cos2x)

= -1/4·xcos2x + 1/4·∫cos2x dx

= -1/4·xcos2x + 1/8·sin2x + C

(5)∫e^(-x)cosxdx

=∫e^(-x)d(sinx)

=e^(-x)·sinx -∫sinx d[e^(-x)]

=e^(-x)·sinx +∫e^(-x)sinx dx

=e^(-x)·sinx -∫e^(-x)d(cosx)

=e^(-x)·sinx - e^(-x)·cosx +∫cosx d[e^(-x)]

=e^(-x)·sinx - e^(-x)·cosx -∫e^(-x)cosxdx