Evaluate: #int_0^2 ln(x+4) dx# ?

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Answer 1 · 2017-05-05T04:42:27

#int_0^2(ln(x+4))dx=6ln6-4ln4-2#

The indefinite integral is:
#int" " ln(x+4)" "dx#

First do a u-substitution (I'll use the letter #w# for clarity, to not get mixed up with integration by parts)
#w=x+4#
#(dw)/dx=1#
#dw=dx#

#int" " lnw " "dw#

Integration by parts:
#((u=lnw), (du=1/w dw))((v=w),(dv=1))#

#wlnw-int(w/w)dw#

#=wlnw-int(1)dw#

#=wlnw-w#
#=(x+4)ln(x+4)-(x+4)#

Definite integral:
#int_0^2(ln(x+4))dx#
#=[(x+4)ln(x+4)-(x+4)]_0^2#

#=[6ln6-6]-[4ln4-4]#

#=6ln6-4ln4-2#

Answer 2 · 2017-05-05T04:49:46

#6ln6-4ln4-2 ~= 3.20538#

#I= int_0^2 ln(x+4) dx#

Let #u=x+4#

#(du)/dx =1 -> du=dx#

#int f(u)g'(u)du = f(u)g(u) - int f'(u)g(u)du#

In this example: #f(u) = lnu# and #g'(u) = 1#

#:. f'(u) = 1/u# and #g(u) = u#

Hence (indefinite) #I= lnu*u - int 1/u *u du#

#= u lnu - int 1du#

#= u lnu - u#

Undo substitution:

#I = [(x+4) ln(x+4) - (x+4)]_0^2#

#= 6ln6-4ln4-6+4#

#=6ln6-4ln4-2 ~= 3.20538#