1-x-1-x-dx-

Question Number 168755 by MikeH last updated on 17/Apr/22

\int \frac{1}{x + \sqrt{1 - x}} d x

Commented by safojontoshtemirov last updated on 17/Apr/22

\int \frac{1}{x + \sqrt{1 - x}} d x = \sqrt{1 - x} = t 1 - x = t^{2} x = 1 - t^{2} d x = - 2 t d t

- \int \frac{2 t}{1 - t^{2} + t} d t = \int \frac{2 t}{t^{2} - t - 1} d t = \int \frac{2 t - 1}{t^{2} - t - 1} d t + \int \frac{1}{t^{2} - t - 1} d t = I_{1} + I_{2}

I_{1} = \int \frac{d (t^{2} - t - 1)}{t^{2} - t - 1} = l n (t^{2} - t - 1) + C_{1} = l n (- x - \sqrt{1 - x}) + C_{1} x ⩽ 0

I_{2} = \int \frac{1}{t^{2} - t - 1} d t = \int \frac{1}{{(t - \frac{1}{2})}^{2} - {(\frac{\sqrt{5}}{2})}^{2}} d t

\int \frac{1}{(t - \frac{1}{2} - \frac{\sqrt{5}}{2}) (t - \frac{1}{2} + \frac{\sqrt{5}}{2})} d t = \frac{1}{\sqrt{5}} \int (\frac{1}{t - \frac{1}{2} - \frac{\sqrt{5}}{2}} - \frac{1}{t - \frac{1}{2} + \frac{\sqrt{5}}{2}}) d t

I_{2} = \frac{1}{\sqrt{5}} l n (t - \frac{1}{2} - \frac{\sqrt{5}}{2}) - \frac{1}{\sqrt{5}} l n (t - \frac{1}{2} + \frac{\sqrt{5}}{2}) + C_{2}

l n (- x - \sqrt{1 - x}) + \frac{1}{\sqrt{5}} l n (\frac{2 \sqrt{1 - x} - 1 - \sqrt{5}}{2 \sqrt{1 - x} - 1 + \sqrt{5}}) + C

Commented by Dildora last updated on 17/Apr/22

a j o y i b

Commented by safojontoshtemirov last updated on 17/Apr/22

t a s h a k u r

Commented by peter frank last updated on 17/Apr/22

thank you

Commented by MikeH last updated on 17/Apr/22

thanks brother

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