dx-x-2-n-x-2-a-

Question Number 123452 by Eric002 last updated on 25/Nov/20

\int \frac{d x}{(x^{2} + n) \sqrt{x^{2} + a}}

Commented by MJS_new last updated on 25/Nov/20

depends on the values of a and n and their

relation \dots

Answered by TANMAY PANACEA last updated on 25/Nov/20

t^{2} = \frac{x^{2} + a}{x^{2} + n}

t^{2} x^{2} + t^{2} n = x^{2} + a

x^{2} (t^{2} - 1) = a - t^{2} n

x^{2} = \frac{a - t^{2} n}{t^{2} - 1}

2 x d x = \frac{(t^{2} - 1) (- 2 t n) - (a - t^{2} n) (2 t)}{{(t^{2} - 1)}^{2}} d t

2 x d x = \frac{- 2 t^{3} n + 2 t n - 2 a t + 2 t^{3} n}{{(t^{2} - 1)}^{2}} d t = \frac{2 t (n - a)}{{(t^{2} - 1)}^{2}} d t

x d x = \frac{t (n - a)}{{(t^{2} - 1)}^{2}} d t

d x = \frac{t (n - a)}{{(t^{2} - 1)}^{2}} \times \frac{1}{x} = \frac{t (n - a)}{{(t^{2} - 1)}^{2}} \times \frac{\sqrt{t^{2} - 1}}{\sqrt{a - t^{2} n}} d t

x^{2} + n

= \frac{a - t^{2} n}{t^{2} - 1} + n = \frac{a - t^{2} n + {n t}^{2} - n}{t^{2} - 1} = \frac{a - n}{t^{2} - 1}

x^{2} + a = \frac{a - t^{2} n}{t^{2} - 1} + a = \frac{a - t^{2} n + {a t}^{2} - a}{t^{2} - 1} = \frac{t^{2} (a - n)}{t^{2} - 1}

\int \frac{d x}{(x^{2} + n) \sqrt{x^{2} + a}}

\int \frac{t (a - n)}{{(t^{2} - 1)}^{2}} \times \sqrt{t^{2} - 1} \times \frac{1}{\sqrt{a - t^{2} n}} \times \frac{t^{2} - 1}{a - n} \times \frac{1}{\sqrt{\frac{t^{2} (a - n)}{t^{2} - 1}}} \times d t

\int \frac{t (a - n)}{{(t^{2} - 1)}^{2}} \times {(t^{2} - 1)}^{2} \times \frac{1}{\sqrt{a - t^{2} n}} \times \frac{1}{{(a - n)}^{\frac{3}{2}}} \times \frac{d t}{t}

\int \frac{1}{{(a - n)}^{\frac{1}{2}}} \times \frac{d t}{\sqrt{n} (\sqrt{\frac{a}{n} - t^{2}})}

\frac{1}{{(a n - n^{2})}^{\frac{1}{2}}} \int \frac{d t}{\sqrt{\frac{a}{n} - t^{2}}}

\frac{1}{{(a n - n^{2})}^{\frac{1}{2}}} {s i n}^{- 1} (\frac{t}{\sqrt{\frac{a}{n}}}) + C

\frac{1}{{(a n - n^{2})}^{\frac{1}{2}}} \times {s i n}^{- 1} (\frac{\sqrt{\frac{x^{2} + a}{x^{2} + n}}}{\sqrt{\frac{a}{n}}}) + C

p l s c h k

Commented by TANMAY PANACEA last updated on 25/Nov/20

p l d c h k e r r o r i f s n y

Commented by MJS_new last updated on 25/Nov/20

what if a n - n^{2} ⩽ 0 or \frac{a}{n} < 0 ?

Answered by nueron last updated on 25/Nov/20

Commented by MJS_new last updated on 25/Nov/20

yeah . but for a = - 4 \land n = 3 you would

probably choose a different path \dots

Answered by TANMAY PANACEA last updated on 25/Nov/20

x = \sqrt{a} t a n θ

d x = \sqrt{a} {s e c}^{2} θ d θ

\int \frac{\sqrt{a} {s e c}^{2} θ d θ}{({a t a n}^{2} θ + n) \sqrt{a} s e c θ}

\int \frac{d θ}{c o s θ (a \frac{{s i n}^{2} θ}{{c o s}^{2} θ} + n)}

\int \frac{c o s θ d θ}{{a s i n}^{2} θ + n (1 - {s i n}^{2} θ)}

\int \frac{d (s i n θ)}{n + (a - n) {s i n}^{2} θ}

\frac{1}{(a - n)} \int \frac{d (s i n θ)}{{(\sqrt{\frac{n}{a - n}})}^{2} + {s i n}^{2} θ}

\frac{1}{(a - n)} \times \frac{\sqrt{a - n}}{\sqrt{n}} {t a n}^{- 1} (\frac{s i n θ}{\sqrt{\frac{n}{a - n}}}) + C

\frac{1}{\sqrt{n} \times \sqrt{a - n}} \times {t a n}^{- 1} (\frac{\frac{x}{\sqrt{a + x^{2}}}}{\sqrt{\frac{n}{a - n}}}) + C

Answered by Bird last updated on 25/Nov/20

I = \int \frac{d x}{(x^{2} + n) \sqrt{x^{2} + a}}

a > 0 w e d o t h e c h a n g e m e n t

x = \sqrt{a} s h (t) \Rightarrow

I = \int \frac{\sqrt{a} c h (t)}{\sqrt{a} c h (t) ({a s h}^{2} t + n)} d t

= \int \frac{d t}{{a s h}^{2} t + n} = \int \frac{d t}{a . \frac{c h (2 t) - 1}{2} + n}

= \int \frac{2 d t}{a c h (2 t) - a + 2 n}

= \int \frac{2 d t}{a \frac{e^{2 t} + e^{- 2 t}}{2} + 2 n - a}

= \int \frac{4 d t}{a (e^{2 t} + e^{- 2 t}) + 4 n - 2 a}

=_{e^{2 t} = u} 4 \int \frac{d u}{2 u {a u + {a u}^{- 1} + 4 n - 2 a}}

= 2 \int \frac{d u}{{a u}^{2} + a + (4 n - 2 a) u}

= 2 \int \frac{d u}{{a u}^{2} + (4 n - 2 a) u + a}

Δ^{^{'}} = {(2 n - a)}^{2} - a^{2}

= 4 n^{2} - 4 n a = 4 n (n - a)

i f n < a \Rightarrow Δ^{^{'}} < 0 \Rightarrow

{a u}^{2} + (4 n - 2 a) u + a

= a {u^{2} + 2 (\frac{2 n - a)}{a}) u + \frac{{(2 n - a)}^{2}}{a^{2}}

+ a - \frac{{(2 n - a)}^{2}}{a^{2}}}

= a {{(u + \frac{2 n - a}{a})}^{2} + \frac{a^{3} - {(2 n - a)}^{2}}{a^{2}}}

i n t h i s c a s e w e d o t h e c h a n g e m e n t

u + \frac{2 n - a}{a} = \sqrt{\frac{a^{3} - {(2 n - a)}^{2}}{a^{2}}} z)

i f n > a Δ^{^{'}} > 0 \Rightarrow u_{1} = \frac{a - 2 n + 2 \sqrt{n^{2} - n a}}{a}

u_{2} = \frac{a - 2 n - 2 \sqrt{n^{2} - n a}}{a}

\Rightarrow I = \frac{2}{a} \int \frac{d u}{(u - u_{1}) (u - u_{2})}

= \frac{2}{a (u_{1} - u_{2})} \int (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u

= \frac{2}{a (u_{1} - u_{2})} l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣ + c

u = e^{2 t} s n d t = a r g s h (\frac{x}{\sqrt{a}})

= l n (\frac{x}{\sqrt{a}} + \sqrt{1 + \frac{x^{2}}{a}})

= l n (\frac{x + \sqrt{a + x^{2}}}{\sqrt{a}}) \Rightarrow

u = {(\frac{x + \sqrt{a + x^{2}}}{\sqrt{a}})}^{2} \dots .

Commented by Bird last updated on 25/Nov/20

i f a < 0 w e d o t h e c h a n g e m e n t

x = \sqrt{- a} c h (t)

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