confusion with log integrals (1 Viewer)

ninetypercent · Jun 8, 2010

i'm confused

I feel like an idiot for asking this but what is the integral of:

$\int \frac{dx}{\sqrt{(x-1)^2 + 9}}$

one textbook says:
$\int \frac{dx}{\sqrt{(x-1)^2 + 9}} = ln\left | x + \sqrt{(x-1)^2 + 9} \right | + C$

another says:
$\int \frac{dx}{\sqrt{(x-1)^2 + 9}} = ln\left | (x-1) + \sqrt{(x-1)^2 + 9} \right | + C$

so whats the actual answer!!!???

have a nice day

Dragonmaster262 · Jun 8, 2010

I think it's it the second one.

lychnobity · Jun 8, 2010

Definitely the second one.

omgmynameislong · Jun 8, 2010

2nd one

Affinity · Jun 8, 2010

ninetypercent said:
i'm confused

I feel like an idiot for asking this but what is the integral of:

$\int \frac{dx}{\sqrt{(x-1)^2 + 9}}$

one textbook says:
$\int \frac{dx}{\sqrt{(x-1)^2 + 9}} = ln\left | x + \sqrt{(x-1)^2 + 9} \right | + C$

another says:
$\int \frac{dx}{\sqrt{(x-1)^2 + 9}} = ln\left | (x-1) + \sqrt{(x-1)^2 + 9} \right | + C$

so whats the actual answer!!!???

have a nice day

try deriving it by using the substitution
$(x - 1 ) = \frac{3}{2}(e^x - e^{-x})$

nat_doc · Jun 8, 2010

second one... the table of SI has x^2... but in this case it is the "x" = x-1

saranghaeyo · Jun 15, 2010

2nd one use u=x-1 to verify

cutemouse · Jun 16, 2010

Affinity said:
try deriving it by using the substitution
$(x - 1 ) = \frac{3}{2}(e^x - e^{-x})$

You don't do hyperbolic functions in 4U...

mirakon · Jun 16, 2010

It's the second one if you actually work it out. Whatever is in the brckets has to be there as well so it's x-1 not just x

confusion with log integrals (1 Viewer)

ninetypercent

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Dragonmaster262

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lychnobity

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omgmynameislong

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Affinity

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nat_doc

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saranghaeyo

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cutemouse

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mirakon

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