两道不定积分计算题?
2个回答
展开全部
2、针对∫f(x)dx,令x=lnt,则dx=dt/t
∫f(x)dx=∫f(lnt)/tdt
=∫ln(1+t)/t^2dt
=-∫ln(1+t)d(1/t)
=-ln(1+t)/t+∫1/t(1+t)dt
=-ln(1+t)/t+∫[1/t-1/(1+t)]dt
=-ln(1+t)/t+ln|t|-ln|1+t|+C,其中C是任意常数
=-ln(1+e^x)/e^x+x-ln(1+e^x)+C
3、∫[lnf(x)+lnf'(x)][f'(x)^2+f(x)f''(x)]dx
=∫[lnf(x)+lnf'(x)]d[f(x)f'(x)]
=[lnf(x)+lnf'(x)]f(x)f'(x)-∫f(x)f'(x)[f'(x)/f(x)+f''(x)/f'(x)]dx
=[lnf(x)+lnf'(x)]f(x)f'(x)-∫[f'(x)^2+f(x)f''(x)]dx
=[lnf(x)+lnf'(x)]f(x)f'(x)-f(x)f'(x)+C,其中C是任意常数
=[lnf(x)+lnf'(x)-1]f(x)f'(x)+C
∫f(x)dx=∫f(lnt)/tdt
=∫ln(1+t)/t^2dt
=-∫ln(1+t)d(1/t)
=-ln(1+t)/t+∫1/t(1+t)dt
=-ln(1+t)/t+∫[1/t-1/(1+t)]dt
=-ln(1+t)/t+ln|t|-ln|1+t|+C,其中C是任意常数
=-ln(1+e^x)/e^x+x-ln(1+e^x)+C
3、∫[lnf(x)+lnf'(x)][f'(x)^2+f(x)f''(x)]dx
=∫[lnf(x)+lnf'(x)]d[f(x)f'(x)]
=[lnf(x)+lnf'(x)]f(x)f'(x)-∫f(x)f'(x)[f'(x)/f(x)+f''(x)/f'(x)]dx
=[lnf(x)+lnf'(x)]f(x)f'(x)-∫[f'(x)^2+f(x)f''(x)]dx
=[lnf(x)+lnf'(x)]f(x)f'(x)-f(x)f'(x)+C,其中C是任意常数
=[lnf(x)+lnf'(x)-1]f(x)f'(x)+C
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询