Question-67907

Question Number 67907 by A8;15: last updated on 02/Sep/19

Commented by mathmax by abdo last updated on 02/Sep/19

l e t I = \int \frac{d x}{x \sqrt{x^{2} + x - 6}}

x^{2} + x - 6 = 0 \to Δ = 1 - 4 (- 6) = 25 \Rightarrow x_{1} = \frac{- 1 + 5}{2} = 2 a n d

x_{2} = \frac{- 1 - 5}{2} = - 3 \Rightarrow I = \int \frac{d x}{x \sqrt{(x - 2) (x + 3)}} w e d o t h e c h a n g e m e n t

\sqrt{x - 2} = t \Rightarrow x - 2 = t^{2} \Rightarrow x = t^{2} + 2 \Rightarrow I = \int \frac{2 t d t}{(t^{2} + 2) t \sqrt{t^{2} + 2 + 3}}

= \int \frac{2 d t}{(t^{2} + 2) \sqrt{t^{2} + 5}} a f t e r w e d o t h e c h a n g e m e n t t = \sqrt{5} s h (u) \Rightarrow

I = \int \frac{2 \sqrt{5} c h (u)}{(25 {s h}^{2} u + 2) \sqrt{5} c h (u)} d u = \int \frac{2 d u}{25 \frac{c h (2 u) - 1}{2} + 2}

= \int \frac{4 d u}{25 c h (2 u) - 25 + 4} = \int \frac{4 d u}{25 c h (2 u) - 21}

= \int \frac{4 d u}{25 \frac{e^{2 u} + e^{- 2 u}}{2} - 21} = \int \frac{4 d u}{25 e^{2 u} + 25 e^{- 2 u} - 42}

=_{e^{2 u} = t} \int \frac{4}{25 t + 25 t^{- 1} - 42} \times \frac{d t}{2 t} = \int \frac{2 d t}{25 t^{2} + 25} = \frac{2}{25} \int \frac{d t}{1 + t^{2}}

= \frac{2}{25} a r c t a n (t) + c = \frac{2}{25} a r c t a n (e^{2 u}) + c

b u t u = a r g s h (\frac{t}{\sqrt{5}}) = l n (\frac{t}{\sqrt{5}} + \sqrt{1 + \frac{t^{2}}{5}}) \Rightarrow

e^{2 u} = {(\frac{t}{\sqrt{5}} + \sqrt{1 + \frac{t^{2}}{5}})}^{2} \Rightarrow I = \frac{2}{25} a r c t a n {{(\frac{t + \sqrt{5 + t^{2}}}{\sqrt{5}})}^{2}} + C

Answered by MJS last updated on 02/Sep/19

\int \frac{d x}{x \sqrt{x^{2} + x - 6}} =

[t = \frac{12 - x}{5 x} \to d x = - \frac{5 x^{2}}{12} d t]

= - \frac{\sqrt{6}}{6} \int \frac{d t}{\sqrt{1 - t^{2}}} = - \frac{\sqrt{6}}{6} \arcsin t =

= \frac{\sqrt{6}}{6} \arcsin \frac{x - 12}{5 x}

Commented by MJS last updated on 02/Sep/19

\int \frac{d x}{x \sqrt{(x + a^{2}) (x - b^{2})}} =

[t = \frac{(a^{2} - b^{2}) x - 2 a^{2} b^{2}}{(a^{2} + b^{2}) x} \to d x = \frac{(a^{2} + b^{2}) x^{2}}{2 a^{2} b^{2}} d t]

= - \frac{1}{\sqrt{a^{2} b^{2}}} \int \frac{d t}{\sqrt{1 - t^{2}}}

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