How do you integrate (x+3)ln^2(x+3)dx?

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Answer 1 · 2018-05-10T08:51:40

#I=(x+3)^2/4[2ln^2(x+3)-2ln(x+3)+1]+c#

Here,

#I=int(x+3)ln^2(x+3)dx#

#I=int(x+3)[ln(x+3)]^2dx#

We take,

#ln(x+3)=t=>x+3=e^t=>dx=e^tdt#

So,

#I=int(e^t)[t]^2e^tdt#

#:.I=intt^2*e^(2t)dt#

#"Using "color(blue)"Integration by parts : "#

#I=t^2inte^(2t)dt-int(d/(dt)(t^2) xx inte^(2t)dt)dt#

#=t^2*(e^(2t))/2-int 2t*(e^(2t))/2dt#

#=t^2*(e^(2t))/2-int t*(e^(2t))dt#

Again, #"using "color(blue)"Integration by parts : "#in second integral,

#I=t^2*(e^(2t))/2-[t*(e^(2t))/2-int(1)(e^(2t)/2)dt]#

#=t^2*(e^(2t))/2-t*e^(2t)/2+1/2inte^(2t)dt#

#=t^2*(e^(2t))/2-t*e^(2t)/2+1/2*e^(2t)/2+c#

#:.I=e^(2t)/4[2t^2-2t+1]+c#

#=>I=(x+3)^2/4[2ln^2(x+3)-2ln(x+3)+1]+c#

Note:

#ln(x+3)=t#

#=>[ln(x+3)]^2=t^2#

#=>ln^2(x+3)=t^2#

Also,

#(i)ln^2(x+3)=[ln(x+3)]^2=ln(x+3)xxln(x+3),and#

#(ii)ln(x+3)^2=2ln(x+3)#

#=>ln^2(x+3)!=ln(x+3)^2#