积分计算问题的求解

∫1/x(x^6+1)dx

=∫(1/[1/t(1/t^6+1)]d(1/t)

=∫(t^7*(-1/t^2)/(1+t^6)dt

=-∫t^5/(1+t^6)dt

=-1/6∫1/(1+t^6)d(1+t^6)

=-1/6ln|1+t^6|+C

代入T=1/x，原公式=-1/6ln | 1+1/x 6 |+c。

2.首先使用乘积和差公式:

sin(A)cos(B)=(1/2)[sin(A+B)+sin(A-B)]

∴sin(2x)cos(3x)=(1/2)[sin(2x+3x)+sin(2x-3x)]

= (1/2)[sin5x + sin(-x)]

= (1/2)(sin5x - sinx)

∴∫ sin(2x)cos(3x) dx

=(1/2)∫sin(5x)dx-(1/2)∫sinx dx

=(1/2)(1/5)∫sin(5x)d(5x)-(1/2)∫sinx dx

=(1/10)(-cos(5x)]+(1/2)cosx+C

=(1/10)[5 cosx-cos(5x)]+C

4.=∫(1/x^4-1/x^6)dx=-1/3*x^(-3)+1/5x^(-5)+c

5.∫(lnx)^3/x^2dx

=-∫(lnx)^3(1/x)'dx

=-(lnx)^3(1/x)+3∫(lnx)^2(1/x)^2dx

=-(lnx)^3(1/x)-3∫(lnx)^2(1/x)'dx

=-(lnx)^3/x-3(lnx)^2/x+6∫lnx(1/x)^2dx

=-(lnx)^3/x-3(lnx)^2/x-6lnx/x+∫1/x^2dx

=-[(lnx)^3+3(lnx)^2+6lnx+1](1/x)+c.