Question

\[\int_{9}^{16} \frac{\sqrt(x)}{x-4} dx\]
\[u=\sqrt(x)\]
\[du=\frac 1 2 \frac{1}{\sqrt(x)}dx\]
\[2\sqrt(x) du=dx\]
\[2udu=dx\]
\[\int_{3}^{4} \frac{u^2}{u^2-4} du\]
\[2\int_{3}^{4} 1+\frac{4}{(u-2)(u+2)} du\]
\[2\int_{3}^{4} du +2\int_{3}^{4}\frac{4}{(u-2)(u+2)} du\]
\[2\int_{3}^{4} du +8\int_{3}^{4} \frac{1}{(u-2)} - \frac{1}{(u+2)}du\]

Answer 1

ok?

Answer 2

Are we on the right track?

Answer 3

yes, there is a slight typo in the third line where the two disappears, but it comes back, so you are fine

Answer 4

ok thank you

Answer 5

\[2\int_{3}^{4} du +8\int_{3}^{4} \frac{1}{(u-2)}du - 8\int_{3}^{4} \frac{1}{(u+2)}du\]

Answer 6

let me check the constants carefully

Answer 7

ok there is a mistake in the constants

Answer 8

I knew it!

Answer 9

\[\frac{4}{(x+2)(x-2)}=\frac{1}{x-2}-\frac{1}{x+2}\] you used the 4 twice it looks like

Answer 10

how so?

Answer 11

how so as in why it is true that
\[\frac{4}{(x-2)(x+2)}=\frac{1}{x-2}-\frac{1}{x+2}\] or how so as in how did you use the 4 twice?

Answer 12

how did i use the 4 twice

Answer 13

\[\frac{4}{(x-2)(x+2)}=\frac{1}{x-2}-\frac{1}{x+2}\]
is the above statement correct though?

Answer 14

yes

Answer 15

so now the 4 is gone right?

Answer 16

yes

Answer 17

and therefore you do not get an 8 out front, but rather the 2 you started with
that is the mistake

Answer 18

Yep

Answer 19

thanks....that 4 looked suspicious

Answer 20

\[2\int_{3}^{4} du +2\int_{3}^{4}\frac{4}{(u-2)(u+2)} du\]
\[=2\int_{3}^{4} du +2\int_{3}^{4} \frac{1}{(u-2)} - \frac{1}{(u+2)}du\]

Answer 21

thanks for catching that!

Answer 22

been at this all day?

Answer 23

yw

Answer 24

no ...i did some other problems in the mean time...but yeah...I kept making small mistakes and confusing myself

Answer 25

that's when I decided to make the face of "confusion" my profile pic

Answer 26

a lot of book keeping
easy to make a mistake

Answer 27

lol