# Answer to Question #23914 in Calculus for Kyle

Question #23914

Can you please help me evaluate this indefinite integral?

∫ [1/(x√[(x^2)-4])dx]

I tried to do it and I got an answer of 1/2 arcsec(x/2) but I know that that's probably wrong because I used a sketchy process to get it to that.

I thought it might just be a special integral that I'm supposed to look up, but I don't have my book with the list of them with me right now and I can't find this specific type of integral listed anywhere on the internet.

Here's what I did:

∫ [1/(x√[(x^2)-4])dx]

∫ dx / (x √[ 4( [(x^2)/4] - 1)])

∫ dx / [2x √( [(x^2)/4] - 1)]

∫ (1/2) * {dx / [ x √( [(x/2)^2] - 1) ] } ....... Let u = x/2, du = 1/2dx, dx=2du

∫ (1/2) * {2du / (x √[ (u^2) - 1] ) }

∫ (1/2) * {du / ( [x/2] √[ (u^2) - 1] ) } ........ Take out constant multiple of 1/2

(1/2) ∫ {du / ( [x/2] √[ (u^2) - 1] ) } .......... *Sketchy Part* I assume (x/2) is equal to |x/2| and then use trig properties to take { } and turn it into arcsec

(1/2) arcsec(u)

(1/2) arcsec(x/2)

Is this somehow right?

∫ [1/(x√[(x^2)-4])dx]

I tried to do it and I got an answer of 1/2 arcsec(x/2) but I know that that's probably wrong because I used a sketchy process to get it to that.

I thought it might just be a special integral that I'm supposed to look up, but I don't have my book with the list of them with me right now and I can't find this specific type of integral listed anywhere on the internet.

Here's what I did:

∫ [1/(x√[(x^2)-4])dx]

∫ dx / (x √[ 4( [(x^2)/4] - 1)])

∫ dx / [2x √( [(x^2)/4] - 1)]

∫ (1/2) * {dx / [ x √( [(x/2)^2] - 1) ] } ....... Let u = x/2, du = 1/2dx, dx=2du

∫ (1/2) * {2du / (x √[ (u^2) - 1] ) }

∫ (1/2) * {du / ( [x/2] √[ (u^2) - 1] ) } ........ Take out constant multiple of 1/2

(1/2) ∫ {du / ( [x/2] √[ (u^2) - 1] ) } .......... *Sketchy Part* I assume (x/2) is equal to |x/2| and then use trig properties to take { } and turn it into arcsec

(1/2) arcsec(u)

(1/2) arcsec(x/2)

Is this somehow right?

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