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

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Answer 1 · 2016-07-12T19:37:41

$= \frac{x^{2} - 1}{2} ln (x + 1) - \frac{1}{4} x (x - 2) + C$ $= (x^2 -1) /2 ln(x+1) - 1/4 x(x - 2) + C$

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

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

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

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

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

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

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

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

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

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