Question

Integrate log { x-6/x-1

Plz help

Answer 1

Like this? \[\int \log\left(\dfrac{x-6}{x-2}\right)dx\]

Answer 2

Yes

Answer 3

Sorry i am on my ipad so cant write proper words

Answer 4

The answer at the back is x-5 ln (x-1) +c

Answer 5

I dunno how to get that

Answer 6

can you check again? I don't think answer is x-5 ln (x-1) + C

Answer 7

The question is Show that x-6/x-1= 1- 5/x-1

Theh it says hence find { x-6/x-1

Answer 8

And answer at the back is x-5 loge (x-1) +c

Answer 9

ah I see, for first part, you have to use polynomial long division.

Answer 10

Or simply split the integral using propertt log a/b=log a-log b

Answer 11

*propery

Answer 12

But how thats gonna help

Answer 13

Because integral of log (x-a)=x(log(x-a)-1)

Answer 14

Do u mind solving it

Answer 15

I don't think I should. I just gave you the formula. Plug in and get your answer.

Answer 16

Honestly i dunno hw that formula apply to fraction

Answer 17

First use log a/b=log a-log b

Then apply formula for each term.

Answer 18

So ∫ x/6 / x/1

Answer 19

i think your question is \[show ~that ~\frac{ x-6 }{ x-1 }=1-\frac{ 5 }{ x-1 } and~hence~integrate~\it.\]

Answer 20

Yes

Answer 21

\[\frac{ x-6 }{ x-1 }=\frac{ x-1-5 }{ x-1 }=\frac{ x-1 }{ x-1 }-\frac{ 5 }{ x-1 }=?\]

Answer 22

I know how to prove the equation its the integrate answer

Answer 23

I dont get where x-5 comes from in x-5 ln (x-1 )+ answer

Answer 24

\[\frac{ x-1 }{ x-1 }-\frac{ 5 }{ x-1 }=1-\frac{ 5 }{ x-1 }\]
\[\int\limits \frac{ x-6 }{ x-1 }dx=\int\limits \left( 1-\frac{ 5 }{ x-1 } \right)dx=\int\limits dx-5 \int\limits \frac{ dx }{ x-1 }+c\]
=?