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

2 Answers
Anjali G
May 5, 2017

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

Explanation:

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

Alan N.
May 5, 2017

6ln6-4ln4-2 ~= 3.20538

Explanation:

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

Let u=x+4

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

Through integration by parts:

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

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