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I'm having a bit of difficulty with this one. Can anyone help?
Many thanks.
Q. log_{e}x dx
Attempt: u = log_{e}x => du = dx
dv = => dv = x^{-2} => v = x^{-2} =>
u dv = uv - v du => log_{e}x dx => log_{e}x + => + => -
log_{e}x dx = => => =>
Ans.: (From text book): (1 - ln2)
I've made the change I think you're referring to but I'm still off the mark...
u = log_{e}x => du = dx
dv = => => v = => ln(x^{2})
u dv = uv - v du => log_{e}x dx => log_{e}x(ln(x^{2})) - ln(x^{2}) => ln(x^{2})log_{e}x - (x^{2}ln(x^{2}) - x^{2})
log_{e}x dx = [ln(x^{2})log_{e}x - (x^{2}ln(x^{2}) - x^{2}) ] => [ln(4)log_{e}2 - (4ln(4) - 4) ] - ln(1)log_{e}1 - (1ln(1) - 1)(1)] => ln(4)log_{e}2 - (4ln(4) - 4) + 0 + 1
Ans.: (From text book): (1 - ln2)
Thanks for that link. Really useful! The only question I have left is with the expression of the answer. The book has it as 1/2(1 - ln2) but the link you provided shows a final answer of 1/2(1 - log2). Is there no difference?
Thanks again, guys.