Question

Evaluation the following integral.

Answer 1

\[\int\limits_{1}^{e}\frac{ 1-lnx }{ x }dx\]

Answer 2

The answer should be 1/2. I just need to see how to do it.

Answer 3

seperate them.
\[\Large \int\limits_{1}^{c} \frac{1}{x}dx-\int\limits_{1}^{c} \frac{\ln(x)}{x}dx\]

Answer 4

can you integrate these ?

Answer 5

you there . can you integrate ?

Answer 6

Um. Well. I know that the integral of 1/x should be lnx.

Answer 7

I mean, you should be helping me with this.

Answer 8

I keep getting an answer that's not 1/2.

Answer 9

the answer is not 1/2 .

Answer 10

One hint for the second integral: think about what the derivative of ln(x) is

Answer 11

You can do a u-substitution with that

Answer 12

\[\int\limits\limits_{1}^{c} \frac{1}{x}dx-\int\limits\limits_{1}^{c} \frac{\ln(x)}{x}dx\]\[=\ln(x)\Large|_1^{c}\small -\int\limits\limits_{1}^{c} \frac{\ln(x)}{x}dx\]\[= \ln(c)-\ln(1) -\int\limits\limits_{1}^{c} \frac{\ln(x)}{x}dx\]
2nd integral:
Let u = ln x, du = 1/x dx

at x = 1, u = ln(1)=0
at x = c, u = ln(c)

\[\int\limits\limits_{0}^{\ln(c)} udu= \]\[\frac{u^2}{2}\Large|_0^{\ln(c)}= \frac{(\ln(c))^2}{2}-0\]

All together: and since ln(1)=0:
\(ln(c)-\frac{(ln(c))^2}{2}\)

Answer 13

Ended up with one half. Thanks.