find-f-x-dt-t-2-xt-1-with-x-real-

Question Number 76356 by mathmax by abdo last updated on 26/Dec/19

f i n d f (x) = \int \frac{d t}{\sqrt{t^{2} - x t + 1}} w i t h x r e a l .

Commented by mathmax by abdo last updated on 28/Dec/19

t^{2} - x t + 1 \to Δ = x^{2} - 4

c a s e 1 ∣ x ∣< 2 \Rightarrow Δ < 0 \Rightarrow t^{2} - x t + 1 = t^{2} - 2 \frac{x}{2} t + \frac{x^{2}}{4} + 1 - \frac{x^{2}}{4}

= {(t - \frac{x}{2})}^{2} + \frac{4 - x^{2}}{4} w e d o t h e c h a n g e m e n t t - \frac{x}{2} = \frac{\sqrt{4 - x^{2}}}{2} u \Rightarrow

f (x) = \int \frac{1}{\sqrt{\frac{4 - x^{2}}{4}} (\sqrt{1 + u^{2}})} \frac{\sqrt{4 - x^{2}}}{2} d u = \int \frac{d u}{\sqrt{1 + u^{2}}}

= l n (u + \sqrt{1 + u^{2}}) + c = l n (\frac{2 t - x}{\sqrt{4 - x^{2}}} + \sqrt{1 + {(\frac{2 t - x}{\sqrt{4 - x^{2}}})}^{2}}) + c

c a s e 2 ∣ x ∣> 2 \Rightarrow Δ > 0 \Rightarrow t_{1} = \frac{x + \sqrt{x^{2} - 4}}{2} a n d t_{2} = \frac{x - \sqrt{x^{2} - 4}}{2}

f (x) = \int \frac{d t}{\sqrt{(t - t_{1}) (t - t_{2})}} [w e d o t h e c h a n g e m e n t \sqrt{t - t_{1}} = u \Rightarrow

t = u^{2} + t_{1} \Rightarrow f (x) = \int \frac{2 u d u}{u \sqrt{u^{2} + t_{1} - t_{2}}}

= 2 \int \frac{d u}{\sqrt{u^{2} + \sqrt{x^{2} - 4}}} =_{u = {(x^{2} - 4)}^{\frac{1}{4}} z} 2 \int \frac{{(x^{2} - 4)}^{\frac{1}{4}} d z}{\sqrt{\sqrt{x^{2} - 4} (z^{2} + 1)}}

= 2 \int \frac{d z}{\sqrt{z^{2} + 1}} = 2 l n (z + \sqrt{1 + z^{2}}) + C

= 2 l n ({(x^{2} - 4)}^{- \frac{1}{4}} u + \sqrt{1 + {(x^{2} - 4)}^{- \frac{1}{2}} u^{2}}) + C

= 2 l n {(x^{2} - 4) \sqrt{t - t_{1}} + \sqrt{1 + {(x^{2} - 4)}^{- \frac{1}{2}} (t - t_{1})})} + C