Menu Close

log-x-2-dx-

Tinku Tara
Pin




Question Number 46 by surabhi last updated on 25/Jan/15
∫(log x)^2 dx
∫(logx)2dx
Answered by surabhi last updated on 04/Nov/14
∫(log x)^2 dx=∫∣(log x)^2 ∙1∣dx =(log x)^2 ∙∫1 dx−∫{(d/dx)(log x)^2 ∙∫1 dx}dx =x(log x)^2 −∫(((2log x)/x)∙x)dx =x(log x)^2 −2∫(log x∙1)dx =x(log x)^2 −2[(log x)∫dx−∫{(d/dx)(log x)∙∫dx}dx] =x(log x)^2 −2[x log x−∫(1/x)∙x dx] =x(log x)^2 −2x log x+2x+C
∫(logx)2dx=∫∣(logx)2⋅1∣dx=(logx)2⋅∫1dx−∫{ddx(logx)2⋅∫1dx}dx=x(logx)2−∫(2logxx⋅x)dx=x(logx)2−2∫(logx⋅1)dx=x(logx)2−2[(logx)∫dx−∫{ddx(logx)⋅∫dx}dx]=x(logx)2−2[xlogx−∫1x⋅xdx]=x(logx)2−2xlogx+2x+C

Pin