dx-x-2-px-x-2-qx-

Question Number 112249 by MJS_new last updated on 07/Sep/20

\int \frac{d x}{(α x^{2} + p x + β) \sqrt{α x^{2} + q x + β}} = ?

Commented by bemath last updated on 07/Sep/20

waw \dots . .

Answered by ajfour last updated on 07/Sep/20

s a y I = \int \frac{d x}{({a x}^{2} + p x + b) \sqrt{{a x}^{2} + q x + b}}

= \int \frac{\frac{d x}{a \sqrt{a}}}{{{(x + \frac{p}{2 a})}^{2} + \frac{b}{a} - \frac{p^{2}}{4 a^{2}}} \sqrt{{(x + \frac{q}{2 a})}^{2} + \frac{b}{a} - \frac{q^{2}}{4 a^{2}}}}

l e t x + \frac{q}{2 a} = \sqrt{(\frac{b}{a} - \frac{q^{2}}{4 a^{2}})} \tan θ = λ \tan θ

I = \frac{1}{a \sqrt{a}} \int \frac{\sec θ d θ}{{(λ \tan θ + \frac{p - q}{2 a})}^{2} + λ^{2} - \frac{(p^{2} - q^{2})}{4 a^{2}}}

= \frac{1}{a \sqrt{a}} \int \frac{\cos θ d θ}{{(λ \sin θ + \frac{(p - q)}{2 a} \cos θ)}^{2} + (λ^{2} - \frac{(p^{2} - q^{2})}{4 a^{2}}) \cos^{2} θ}

= \frac{1}{a \sqrt{a}} \int \frac{\cos θ d θ}{λ^{2} - \frac{q (p - q)}{2 a^{2}} \cos^{2} θ + \frac{λ (p - q)}{2 a} \sin 2 θ}

l e t θ = \frac{π}{4} - ϕ

I = \frac{1}{a \sqrt{a}} \int \frac{- \cos (\frac{π}{4} - ϕ) d ϕ}{λ^{2} - \frac{q (p - q)}{4 a^{2}} - \frac{(p - q)}{4 a^{2}} {q \sin 2 ϕ - 2 λ a \cos 2 ϕ}}

n o w l e t \sqrt{q^{2} + 4 λ^{2} a^{2}} = h = \sqrt{4 a b}

\tan δ = \frac{q}{2 λ a}; λ^{2} - \frac{q (p - q)}{4 a^{2}} = k = \frac{b}{a} - \frac{p q}{4 a^{2}}

I = \frac{1}{a \sqrt{a}} \int \frac{- \cos (\frac{π}{4} - ϕ) d ϕ}{k + \frac{(p - q) h}{4 a^{2}} \cos (2 ϕ + δ)}

= \frac{1}{a \sqrt{a}} \int \frac{- \cos (ϕ + \frac{δ}{2} - \frac{π}{4} - \frac{δ}{2}) d (ϕ + \frac{δ}{2})}{k + \frac{(p - q) h}{4 a^{2}} {\cos^{2} (ϕ + \frac{δ}{2}) - \sin^{2} (ϕ + \frac{δ}{2})}}

n o w s a y ϕ + \frac{δ}{2} = z, \frac{π}{4} + \frac{δ}{2} = β, t h e n

\frac{h (q - p)}{4 \sqrt{a}} I = \cos β \int \frac{\cos z d z}{\frac{4 a^{2} k}{p - q} + 1 - 2 \sin^{2} z}

+ \sin β \int \frac{\sin z d z}{\frac{4 a^{2} k}{p - q} - 1 + 2 \cos^{2} z}

\dots \dots .

Commented by MJS_new last updated on 07/Sep/20

thank you!

Answered by MJS_new last updated on 07/Sep/20

Mr . Mathdave showed us another path . It

works with \int \frac{d x}{({a x}^{2} + b x + c) \sqrt{{d x}^{2} + e x + f}} only

if a = d \land c = f

\int \frac{d x}{(α x^{2} + p x + β) \sqrt{α x^{2} + q x + β}} =

= α^{- 3 / 2} \int \frac{d x}{(x^{2} + \frac{p}{α} x + \frac{β}{α}) \sqrt{x^{2} + \frac{q}{α} x + \frac{β}{α}}} =

[t = \frac{\sqrt{β} - x \sqrt{α}}{\sqrt{β} + x \sqrt{α}} \to d x = - \frac{2 \sqrt{β}}{{(t + 1)}^{2} \sqrt{α}} d t]

= 2 α^{7 / 4} \int \frac{t + 1}{((p - 2 \sqrt{α β}) t^{2} - (p + 2 \sqrt{α β})) \sqrt{(2 \sqrt{α β} - q) \sqrt{β} t^{2} + (2 \sqrt{α β} + q) \sqrt{β}}} d t =

[a = p - 2 \sqrt{α β} b = - (p + 2 \sqrt{α β})

c = (2 \sqrt{α β} - q) \sqrt{β} d = (2 \sqrt{α β} + q) \sqrt{β}]

= 2 α^{7 / 4} \int \frac{t + 1}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t =

= 2 α^{7 / 4} (\int \frac{t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t + \int \frac{d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}})

both are easy to solve :

\int \frac{t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} d t =

[u = \frac{\sqrt{d}}{\sqrt{{c t}^{2} + d}} \to d t = - \frac{{({c t}^{2} + d)}^{3 / 2}}{c t \sqrt{d}} d u]

= \sqrt{d} \int \frac{d u}{(a d - b c) u^{2} - a d} = \dots

\int \frac{d t}{({a t}^{2} + b) \sqrt{{c t}^{2} + d}} =

[u = \frac{t \sqrt{a d - b c}}{\sqrt{{c t}^{2} + d}} \to d t = \frac{{({c t}^{2} + d)}^{3 / 2}}{d \sqrt{a d - b c}} d u]

= \frac{1}{\sqrt{a d - b c}} \int \frac{d u}{u^{2} + b} = \dots

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