Question

integrate 1/(x^3(sqrt(x^2-1))

Answer 1

here is the deal you com and help me on mkine and I will help u on yrs deal or no deal/

Answer 2

\[\int\limits_{}^{}\frac{ 1 }{ x ^{3}(\sqrt{x ^{2}-1}) }\]

Answer 3

i help with wat?

Answer 4

substitute.

x=sec(u)
dx = tan(u)sec(u) du

and then,

√(x^2-1)=√(sec^2u-1)=√tan^2u = tan u
u= sec^(-1)x

Answer 5

ok let me see. I did get somewhere , let me write where i got to

Answer 6

sure

Answer 7

(I wondered if I can help with integration before I actually even learn derivatives in class)

Answer 8

you are patient :) writing it out. But just saying, that no matter what you get, derive it to check.

Answer 9

\[x =\sec theta\]
\[dx=\sec theta tan theta d theta\]
\[\int\limits_{}^{}\frac{ \sec \theta \tan \theta }{\sec ^{3}\theta (\sqrt{\tan ^{2}\theta}) }\]
\[\int\limits_{}{\frac{ 1 }{ \sec ^{2}\theta }}\]
\[\int\limits_{}^{}\cos ^{2}\theta\]
\[= (1/2 )(\theta + \sin \theta \cos \theta)\]
\[\frac{ 1 }{ 2}\sec ^{-1}x+ what?\]

Answer 10

i dont know how to make 2sin@cos@ to be\[\frac{ \sqrt{x ^{2}-1} }{ 2x }\]

Answer 11

sorry its \[\frac{ 1 }{ 2 }\sin \theta \cos \theta\]

Answer 12

well cos^2u is correct.

then re-write cos^2u as 1/2cos(2u)+1/2

Answer 13

i dont know how to make it be in fraction form, if my working is alryt. I have cut some stages short

Answer 14

thats what i did for the integral and i got\[1/2(\theta +\sin \theta \cos \theta)\]

Answer 15

\[x=\sec \theta\]
\[\theta=\sec ^{-1}x\]
\[then \frac{ 1 }{ 2 }\theta = makes \theta =2\sec ^{-1}x\]

Answer 16

I am getting

S 1/2cos(2u) +1/2 du
S 1/2cos(2u) du + S 1/2 du
sin(2u)/4 + u/2

Answer 17

and final answer then is

1/2 (x^2) / √(x^2-1) - (1/2) tan^(-1) (1/ √(x^2-1)) + C

Answer 18

uhmmm, let me retrace your steps. and pliz just hint me on the following
\[\int\limits_{}^{}\frac{ 1 }{ 1+2e ^{x}+e ^{-x} }\].
What type of substitution or trgi should I make use of? Partial fractions or some factorisation?

Answer 19

this is what I would do. u=e^x

Answer 20

ok, let me c

Answer 21

I think I have to go right now -:(
Well... don't worry about me, I am not such a great helper anyways... bye

Answer 22

thanx man. Just that the u substitution u opted for is a dark tunnel to me. i dont see the light