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

S xln(1+x)dx

## 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

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tiny-tim
Homework Helper
hi vande060!

(have an integral: ∫ and try using the X2 icon just above the Reply box )
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 they're the same!

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

hi vande060!

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

erm they're the same!

(except your -(x+1) should be +(x+1))
thanks for the conformation, the algebra got the best of me this time