Did i solve this integral correctly: S xln(1+x)dx

  • Thread starter vande060
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  • #1
vande060
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Homework Statement



S xln(1+x)dx



Homework Equations





The Attempt at a Solution



S xln(1+x)dx

t = (1+x)
dt = dx
x = t-1

S (t-1)ln(t)dt

u = ln(t)
du = 1/t
dv = (t-1)dt
v =( t^2/2 - t)

uv - S vdu

ln(t)*( t^2/2 - t) - S ( t^2/2 - t)*1/t dt

distributing and solving integral

ln(t)*t^2/2 - ln(t)*t - (t^2/4 - t)

now putting the expression for t back in gives me something radically different from what the book tells me it is

my answer ln(1+x)*[(1+x)^2]/2 - ln(x+1)*(x+1) - [(1+x)^2]/4 - (x+1) + C

the books answer : 1/2*(x^2 - 1)*ln(x+1) - 1/4*x^2 + 1/2*x +3/4 + C


where have i gone wrong
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi vande060! :smile:

(have an integral: ∫ and try using the X2 icon just above the Reply box :wink:)
my answer ln(1+x)*[(1+x)^2]/2 - ln(x+1)*(x+1) - [(1+x)^2]/4 - (x+1) + C

the books answer : 1/2*(x^2 - 1)*ln(x+1) - 1/4*x^2 + 1/2*x +3/4 + C

erm :redface:they're the same! :biggrin:

(except your -(x+1) should be +(x+1))
 
  • #3
vande060
186
0
hi vande060! :smile:

(have an integral: ∫ and try using the X2 icon just above the Reply box :wink:)


erm :redface:they're the same! :biggrin:

(except your -(x+1) should be +(x+1))

thanks for the conformation, the algebra got the best of me this time
 

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