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Mathematics
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Integration by Parts
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\[xln(x+1) dx\]
any ideas?
it's probably best to choose u=x and ln(x+1)dx=dv. The integral for the log will have to be an integration by parts within itself where you choose s=ln(x+a) and dt=dx
that should say ln(x+1) not x+a, typo
I got u = ln(x+1) and du = 1/(x+1). Dv = x and v = x^2
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i dont know, this one has me confused
No don't do it that way, do it the other way around
pick u=x and dv=ln(x+1)dx
Ok, so u = x ; du = 1. dv = ln(x+1) dx ; v = xln(x+1)-x+ln|x+1|
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