Question

【如图】这四道求不定积分的题目怎么用分部积分法求出来？

Answer 1

(1)
∫ln(x^2+1) dx
=xln(x^2+1) - 2∫x^2/(x^2+1) dx
=xln(x^2+1) - 2∫[ 1-1/(x^2+1)] dx
=xln(x^2+1) - 2x +2arctanx +C
(2)
∫ln(lnx)/x dx
=∫ln(lnx) dlnx
=lnx .ln(lnx) - ∫ dx/x
=lnx .ln(lnx) - ln|x| +C
(3)
∫x/(cosx)^2 dx
=∫x(secx)^2 dx
=∫x dtanx
=xtanx - ∫tanx dx
=xtanx + ln|cosx| +C
(4)

∫(1/x^3) e^(1/x) dx
=-∫(1/x) de^(1/x)
=-(1/x)e^(1/x) -∫(1/x^2) e^(1/x) dx
=-(1/x)e^(1/x) + e^(1/x) +C

追问

麻烦请问一下这几道怎么做

Answer 2

1. ∫ ln(x² + 1) dx
= xln(x² + 1) - ∫ x dln(x² + 1)
= xln(x² + 1) - ∫ x · (2x)/(x² + 1) dx
= xln(x² + 1) - 2∫ x²/(x² + 1) dx
= xln(x² + 1) - 2∫ [(x² + 1) - 1]/(x² + 1) dx
= xln(x² + 1) - 2∫ [1 - 1/(x² + 1)] dx
= xln(x² + 1) - 2(x - arctan(x)) + C
= xln(x² + 1) - 2x + 2arctan(x) + C

追答

3. ∫ (x/cos²x) dx
= ∫ x dtanx
= x tanx - ∫ tanx dx + c
= x tanx + ∫ (dcosx)/cosx + c
= x tanx + ln |cosx| + c

3. 先设t=1/x,则 x=1/t 
上式可变为 
∫t^3e^t d(1/t)
=∫ - te^t dt
= - te^t -∫ [e^t d(-t)]
=- te^t+ e^t 
=(1-t)e^t 
在把t=1/x带入,有
∫(1/x^3)*e^(1/x) dx=（1 - 1/x)e^(1/x)

追问

麻烦请问一下这几道怎么做