Question

Integrate the function.

dx/(x(sqrt 9x^2-25))

Answer 1

No, that doesn't work. Try this substitution, x = 5/3 sec u

Answer 2

i have a question as to why it is 5/3 sec U

Answer 3

because we want to deal with all of that messiness inside the square root, and this substitution will make it easy.

in particular, sec because sec^2 u - 1 = tan^2 u
and 5/3 because that deals with all the constants.

Answer 4

are you sure its not 3/5

Answer 5

because if i do it your way the answer is...... 5/3 sec^-1 (5/3x) +c

Answer 6

No 5/3. We want to be able to write that as
\[ \sqrt{k^2(\sec^2 u - 1)} \]
and 5/3 enables us to cancel out the 9 and generate the 25

Answer 7

ok so then what?

Answer 8

right ... so work through it, as it seems you already substantially have.

Answer 9

so its 1/5 sec^-1 (3/5x)+c?

Answer 10

looks about right. arcsec is a slightly bizarre function; if you're feeling motivated, it's better to rewrite it as arctan

Answer 11

ok...not too ambitious here haha

Answer 12

thank you though!

Answer 13

sure

Answer 14

\[\Large \frac{1}{x \sqrt{9
x^2-25}}=\frac{9 x}{x^2
\sqrt{9 x^2-25}}=\\ \Large\frac{9
x}{\left(9
x^2-25\right)^{3/2}
\left(\frac{5^2}{(\sqrt{9
x^2-25})^2}+1\right)}\\

\Large u=\frac{5}{\sqrt{9
x^2-25}}

\]
The rest is data processing.

Answer 15

\[
-\frac{1}{5} \tan
^{-1}\left(\frac{5}{\sqrt{9
x^2-25}}\right) +C
\]
should be your answer.