lets see here, im trying to integrate this(and sorry, i dont know how to use the symbols - ill use '{' as my integral sign):(adsbygoogle = window.adsbygoogle || []).push({});

6

{ 1 / (t^2 - 9) ^ .5

4

so, considering { 1 / (x^2 - 1) ^ .5 = inverse cosh(x) i did:

6

(1/3) { 1 / ((t / 3) ^ 2 - 1) ^ .5

4

then i took the integral so i got:

6

(1/3) inverse cosh(t / 3) |

4

so then, considering inverse cosh(u) = ln(u + (u - 1) ^ .5)

i made my equation into:

6

(1/3) ln((t/3) + ((t/3) - 1) ^ .5)) |

4

and then computing that i come up with .1738641452 but the books answer is :

ln ((6 + 3(3)^.5)/(4 + 7^.5)) which is not the same as mine(plus it is 'prettier' in the sense that it looks like i could get that answer without a calculator which is how im supposed to be doing it).

any help or suggestions? sorry about the readability

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# Homework Help: Inverse hyperbolic function integral

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