quick integration q (1 Viewer)

[onebytwo]

can someone plz help integrate;

3
S 1/rt(x^2-1)dx
rt2

ie from root 2 to 3

thanks for any help

 

[pLuvia]

int{rt2 to 3}1/sqrt{x²-1}dx
=[ln|x-sqrt{x²-1}|]{rt2 to 3}
=[ln(sqrt2-1)/(3-sqrt2)]
=[ln(2sqrt2-1)/7]

 

[hyparzero]

int{rt2 to 3}1/sqrt{x²-1}dx
=[ln|x-sqrt{x²-1}|]{rt2 to 3}
=[ln(sqrt2-1)/(3-sqrt2)]
=[ln(2sqrt2-1)/7]

Shouldn't =[ln|x-sqrt{x²-1}|]{rt2 to 3}

actually be

=[ln|x+sqrt{x²-1}|]{rt2 to 3}

its supposed to me a "plus" after the first x? Unless i'm wrong...

 

[pLuvia]

Typo

 

[onebytwo]

???????
the answer is supposed to be ln(1+rt2)

 

[Riviet]

In the substitution line, you should get:

ln[(3+sqrt8)/(sqrt2+1)]

=ln[(3+sqrt8)(sqrt2+1)] by rationalising the denominator

=ln(3sqrt2 - 3 + sqrt16 - 2sqrt2)

=ln(1+sqrt2)

 

[hyparzero]

???????
the answer is supposed to be ln(1+rt2)

pLuvia made a mistake in the calculation due to the incorrect integration result.

The correct outcome should be:
ln[x+sqrt(x2-1)] from {rt2 to 3}

=> ln[ 3+ sqrt(9 - 1)] - ln[rt2 + sqrt(2 - 1)]
=> ln[3 + sqrt(8)] - ln[rt2 + 1]
=> ln[3 + 2rt2] - ln[rt2 + 1] (log manipulation)

=> ln[(3 + 2rt2) / (rt2 + 1)]
=> ln(1 + rt2)

Its easy to slip up with silly mistakes...

 

[onebytwo]

thanks for the help guys