Is there a different approach to evaluate ∫ln⁡(x)dx?

tebollahb

tebollahb

Answered

2022-01-21

Is there a different approach to evaluate ln(x)dx?

Answer & Explanation

Melinda McCombs

Melinda McCombs

Expert

2022-01-21Added 38 answers

lnx=n=1(1)n+11n(x1)n
lnxdx=n=1(1)n+11n11+n(x1)n+1
=n=1(1)n+1(1n1n+1)(x1)n+1
=(x1)n=1(1)n+11n(x1)n+n=1(1)n+21n+1(x1)n+1
=(x1)lnx+lnx(x1)
=xlnxx+C
Nadine Salcido

Nadine Salcido

Expert

2022-01-22Added 34 answers

Im
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

For fun, a bit of trickery: Let x=ey. Then: lnxdx=yeydy; Now look at : ddy(yey)=ey+yey. Thus: yeydy= ddy(yey)eydy= yeyey=xlnxx+C.

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