How do you integrate #x ln (x + 1) dx#?

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Answer 1 · 2016-07-12T19:37:41

#= (x^2 -1) /2 ln(x+1) - 1/4 x(x - 2) + C#

#int \ x ln (x + 1) \ dx#

#= int \ d/dx (x^2/2) ln (x + 1) \ dx#

and by IBP : #int u v' = uv - int u'v#

#= x^2 /2 ln(x+1) - int \ x^2/2 d/dx ( ln (x + 1) )\ dx#

#= x^2 /2 ln(x+1) - 1/2 color{blue}{ int \ (x^2)/(x + 1) }\ dx qquad triangle#

for the integration in blue, we use a simple sub

#u = x+1, du = dx#

# int \ ((u-1)^2)/(u) \ du#

# int \ u - 2 + 1/u \ du#

# = u^2/2 - 2 u + ln u#

# = (x+1)^2/2 - 2(x+1) + ln (x+1)#

so #triangle# becomes

#= x^2 /2 ln(x+1) - 1/2 ((x+1)^2/2 - 2(x+1) + ln (x+1)) + C#

#= x^2 /2 ln(x+1) - 1/4 (x+1)^2 + (x+1) - 1/2ln (x+1) + C#

#= (x^2 -1) /2 ln(x+1) - 1/4 (x^2 + 2x +1 - 4x-4) + C#

#= (x^2 -1) /2 ln(x+1) - 1/4 (x^2 - 2x - 3) + C#

#= (x^2 -1) /2 ln(x+1) - 1/4 (x^2 - 2x) + C#

#= (x^2 -1) /2 ln(x+1) - 1/4 x(x - 2) + C#