Question

积分计算题求解

谢谢大佬~还有一道 ∫(lnx)^3／x^2dx，这几个是转不过来弯的那种，最近做题做傻了……

Answer 1

1.设x=1/t

∫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

将t=1/x代入,原式=-1/6ln|1+1/x^6|+c

2.先利用积化和差公式:

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)[5cosx - 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.

追问

那个。第三题能做吗

追答

原式=∫x/(1+cosX)dx+∫sinX/(1+cosX)dx
=∫xsec^2(x/2)d(x/2)-∫1/(1+cosx)d(1+cosx)
=∫xd[tan(x/2)]-ln(1+cosx)
=xtan(x/2)-∫tan(x/2)dx-ln(1+cosx)
=xtan(x/2)+2ln∣cos(x/2)∣-ln2-ln∣cos(x/2)∣+C1
=xtan(x/2)+C

追问

十分感谢

Answer 2

嗯这个题确实不是很好做的哈，不是很懂的哈，不是很好的