integrate logx?

Answer 1

well another technique is integration by parts. Let u(x) and v(x) be differentiable on a certain interval. We would like to integrate the function u×v′, but we don't know how. By the product rule, the derivative of uv is uv′+vu′. Thus the integral of uv′ + the integral of vu′ yields uv. Write it this way.
∫ uv′ = uv - ∫ vu′
We may know how to derive the latter integral, thus giving the former.
Assume we want to integrate log(x). (Log is defined in the next section; its derivative is 1/x.) Let u = log(x) and let v = x. Thus log(x) is uv.By parts, the integral becomes x*log(x) - the integral of (x times the derivative of log(x)). The second integrand reduces to 1, hence the answer is x*log(x)-x. Take the derivative to reproduce log(x).
so if f(x) = Log(x)
∫log(x) = x Log(x) - x

I hope i could give you prefect answer.
good luck & good question.

Answer 2

Log X Integration

Answer 3

Integration Of Log X

Answer 4

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RE:
integrate logx?

Answer 5

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Answer 6

int(log x)=
int(1.logx)
now integrating by parts,
=
(int(1dx)*logx) - int(int(1dx)*d(logx)dx)
=
x*log x - int(x/x)
=
x*log x - x+C
=
x*(log x -1)+C

Answer 7

Integral log(x)dx = x*log(x) - x + C if log = natural log
integral log(x)ds = x*log(e)*log(x) - log(e)*x + C if log = base 10 log

EDITED

Answer 8

I think it is ( x-log x ).

and derivative of log x = 1/x

Answer 9

If it is integral log(x).dx then answer is already there [i.e. xlogx-x] but if not dx then you must specify

Answer 10

xlogx-x+c