Question

integrate 1/x^2-25. Evaluate from 6 to 7

Answer 1

I'm doing fraction decomp

Answer 2

so \[1=\frac{ A }{ x+5 }+\frac{ B }{x-5 }\]

Answer 3

I got A=-1/10 and B=1/10

Answer 4

\[\int\limits_{6}^{7}(\frac{- 1 }{ 10 })(\frac{ 1 }{ x+5 })+\int\limits_{6}^{7}(\frac{ 1 }{ 10 })(\frac{ 1 }{ x-5 })\]

Answer 5

then you would have\[\frac{A}{x+5}+\frac{B}{x-5}\\if\space A = \frac{-1}{10
}\\then\\\frac{\frac{-1}{10}}{x+5}+\frac{\frac{1}{10}}{x-5}=\frac{1}{10}[\frac{1}{x-5}-\frac{1}{x+5}]\]

Answer 6

so you have
\[\frac{1}{10}\int_{6}^{7}\frac{1}{x-5}dx-\frac{1}{10}\int_{6}^{7}\frac{1}{x+5}dx\]

Answer 7

\[\frac{ 1 }{ 10 }\ln (x-5)-\frac{ 1 }{ 10 }\ln (x+5)\]

Answer 8

from 6 to 7

Answer 9

So \[\frac{ 1 }{ 10 }\ln(11)-\left[ \frac{ 1 }{ 10 }\ln(2)-\frac{ 1 }{ 10 }\ln(12) \right]\]

Answer 10

correct

Answer 11

Awesome! Thank you :) I think I was getting my signs mixed up

Answer 12

:) np