Integration (1 Viewer)

MC Squidge · May 30, 2009

Integral of ((x^2)((x-1)^(1/2)))^-1 with respect to x

kaz1 · May 30, 2009

Try using the substitution u²=x-1.

study-freak · May 30, 2009

kaz1 said:
Try using the substitution u²=x-1.

where u>_0

jet · May 30, 2009

Drongoski · May 30, 2009

jetblack2007 said:
$\int \frac{dx}{x^2 \sqrt{x - 1}} \\ \text{Let } x = sec^2u \\ dx = 2sec^2u \, tanu \, du \\ \int \frac{dx}{x^2 \sqrt{x - 1}} = \int \frac{2sec^2 u \, tanu \, du}{sec^4 u\sqrt{sec^2 - 1}} \\ = 2\int \frac{du}{sec^2 u} \\ = 2\int cos^2u \, du \\ = \left [u + \frac{1}{2}sin2u \right ] + C \\ = u + sinu \, cosu + C \\ \text{Now, from the substitution, } cosu = \frac{1}{\sqrt{x}}, \therefore sinx = \frac{\sqrt{x -1}}{\sqrt{x}}, \text{ and } tan^{-1}(\sqrt{x - 1}) = u \\ = tan^{-1}\sqrt{x -1} + \frac{\sqrt{x - 1}}{x} + C$

You are a genius !

jet · May 30, 2009

Drongoski said:
You are a genius !

Thanks

MC Squidge · May 30, 2009

yea wow thanks bro

Integration (1 Viewer)

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