find-1-dx-x-2-2-x-3-

Question Number 38104 by maxmathsup by imad last updated on 21/Jun/18

f i n d \int_{1}^{+ \infty} \frac{d x}{(x^{2} + 2) \sqrt{x + 3}}

Answered by MJS last updated on 23/Jun/18

\int \frac{d x}{(x^{2} + 2) \sqrt{x + 3}} =

[t = \sqrt{x + 3} \to d x = 2 \sqrt{x + 3} d t]

= 2 \int \frac{d t}{t^{4} - 6 t^{2} + 11} =

[t^{4} - 6 t^{2} + 11 = 0 \Rightarrow

\Rightarrow t_{1} = - \frac{\sqrt{2}}{2} ((3 + \sqrt{11}) + (- 3 + \sqrt{11}) i)

t_{2} = - \frac{\sqrt{2}}{2} ((3 + \sqrt{11}) - (- 3 + \sqrt{11}) i)

t_{3} = \frac{\sqrt{2}}{2} ((3 + \sqrt{11}) + (- 3 + \sqrt{11}) i)

t_{4} = \frac{\sqrt{2}}{2} ((3 + \sqrt{11}) - (- 3 + \sqrt{11}) i)

(t - t_{1}) (t - t_{2}) = t^{2} + \sqrt{6 + 2 \sqrt{11}} t + \sqrt{11}

(t - t_{3}) (t - t_{4}) = t^{2} - \sqrt{6 + 2 \sqrt{11}} t + \sqrt{11}]

= 2 \int \frac{d t}{(t^{2} - \sqrt{6 + 2 \sqrt{11}} t + \sqrt{11}) (t^{2} + \sqrt{6 + 2 \sqrt{11}} t + \sqrt{11})} =

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

= 2 (\int \frac{P_{1}}{t^{2} - a t + b} d t + \int \frac{P_{2}}{t^{2} + a t + b} d t) =

[this might be hard to find :

we need P_{1} (t^{2} + a t + b) + P_{2} (t^{2} - a t + b) = 1

set P_{1} = a - t \land P_{2} = a + t \Rightarrow

\Rightarrow (a - t) (t^{2} + a t + b) + (a + t) (t^{2} - a t + b) = 2 a b \Rightarrow

\Rightarrow P_{1} = \frac{a - t}{2 a b}; P_{2} = \frac{a + t}{2 a b}]

= \frac{1}{a b} (- \int \frac{t - a}{t^{2} - a t + b} d t + \int \frac{t + a}{t^{2} + a t + b} d t) =

[t - a = \frac{2 t - a}{2} - \frac{a}{2}; t + a = \frac{2 t + a}{2} + \frac{a}{2}]

= \frac{1}{2 a b} (- \int \frac{2 t - a}{t^{2} - a t + b} d t + a \int \frac{d t}{t^{2} - a t + b} + \int \frac{2 t + a}{t^{2} + a t + b} d t + a \int \frac{d t}{t^{2} + a t + b}) =

= \frac{1}{2 a b} (\int \frac{2 t + a}{t^{2} + a t + b} d t - \int \frac{2 t - a}{t^{2} - a t + b} d t) + \frac{1}{2 b} (\int \frac{d t}{t^{2} - a t + b} + \int \frac{d t}{t^{2} + a t + b}) =

= \frac{1}{2 a b} \ln ∣ \frac{t^{2} + a t + b}{t^{2} - a t + b} ∣ + \frac{1}{2 b} (\int \frac{d t}{{(t - \frac{a}{2})}^{2} + b - \frac{a^{2}}{4}} + \int \frac{d t}{{(t + \frac{a}{2})}^{2} + b - \frac{a^{2}}{4}}) =

[t = u \sqrt{b - \frac{a^{2}}{4}} \pm \frac{a}{2} \Rightarrow u = \frac{t \pm \frac{a}{2}}{\sqrt{b - \frac{a^{2}}{4}}} \to d t = \sqrt{b - \frac{a^{2}}{4}} d u]

= \frac{1}{2 a b} \ln ∣ \frac{t^{2} + a t + b}{t^{2} - a t + b} ∣ + \frac{\sqrt{4 b - a^{2}}}{4 b} (\int \frac{{d u}_{1}}{u_{1}^{2} + 1} + \int \frac{{d u}_{2}}{u_{2}^{2} + 1}) =

= \frac{1}{2 a b} \ln ∣ \frac{t^{2} + a t + b}{t^{2} - a t + b} ∣ + \frac{\sqrt{4 b - a^{2}}}{4 b} (\tan^{- 1} u_{1} + \tan^{- 1} u_{2}) =

= \frac{1}{2 a b} \ln ∣ \frac{t^{2} + a t + b}{t^{2} - a t + b} ∣ + \frac{\sqrt{4 b - a^{2}}}{4 b} (\tan^{- 1} (\frac{\sqrt{4 b - a^{2}}}{2} (t - \frac{a}{2})) + \tan^{- 1} (\frac{\sqrt{4 b - a^{2}}}{2} (t + \frac{a}{2}))) =

= \frac{\sqrt{11}}{44} \sqrt{- 3 + \sqrt{11}} \ln ∣ \frac{x + \sqrt{2 (3 + \sqrt{11}) (x + 3)} + 3 + \sqrt{11}}{x - \sqrt{2 (3 + \sqrt{11}) (x + 3)} + 3 + \sqrt{11}} ∣ +

+ \frac{\sqrt{22}}{44} \sqrt{- 3 + \sqrt{11}} (\tan^{- 1} (\frac{\sqrt{2}}{2} (\sqrt{(- 3 + \sqrt{11}) (x + 3)} - 1) + \tan^{- 1} (\frac{\sqrt{2}}{2} (\sqrt{(- 3 + \sqrt{11}) (x + 3)} + 1)))) + C

Commented by abdo mathsup 649 cc last updated on 23/Jun/18

t h a n k y o u s i r M j s .

Commented by MJS last updated on 23/Jun/18

{you}^{'} re welcome as always .

I^{'} m integral - addicted \dots

I {didn}^{'} t solve it in [1; \infty [because {it}^{'} s got no

“ {nice}^{″} solution due to the logarithmic part

but {it}^{'} s \approx . 261683

Commented by MJS last updated on 23/Jun/18

but in [- 3; \infty [{it}^{'} s \frac{\sqrt{22}}{44} \sqrt{- 3 + \sqrt{11}} π

Commented by math khazana by abdo last updated on 23/Jun/18

t h a n k y o u s i r \dots