Question

integrate (sqrt((x-1)/x^5)) using substitution (no trig substitution)

Answer 1

\[\int\limits_{}^{}\sqrt{\frac{x-1}{x^5}} dx\]

Answer 2

this one is really..........not that tough

Answer 3

take out sqrt (1/x^4)

Answer 4

by "take out" i mean we can rewrite the problem as:

\[\int\limits_{}^{}\sqrt{1/x^4}* \sqrt{x-1/x}\]

Answer 5

clever

Answer 6

typo though right?

Answer 7

\[\int\frac{1}{x^2}\sqrt{1-\frac{1}{x}}dx\]

Answer 8

i posted one for you myininaya
just in case you have to teach cacl 1 instead of calc 2

Answer 9

\[\int\limits_{}^{}\frac{1}{x^2} \sqrt{\frac{x-1}{x}} dx\]
let u=x-1
du=dx

if u=x-1=> u+1=x
so we have
\[\int\limits_{}^{}\frac{1}{(u+1)^2} \cdot \sqrt{\frac{u}{u+1}} du\]

let s=u/(u+1)
then ds=1/(u+1)^2 du
so we have
\[\int\limits_{}^{}\sqrt{s} ds\]
gj lagrange

Answer 10

what is wrong with
\[u=1-\frac{1}{x}\]
\[du=\frac{1}{x^2}dx\] get
\[\int\sqrt{u}du\]?

Answer 11

thats the way i would have done it :)

Answer 12

nothing wrong with it

Answer 13

lol
i do my own thing

Answer 14

leave me alone

Answer 15

more than one way to skin a muskrat. i am waiting for you to prove the product rule from the rule for squares

Answer 16

ok zarkon will probably answer before me but i will take a stab at it