prove-that-0-3-2-x-x-2-2x-5-dx-4-13049-

Question Number 98679 by M±th+et+s last updated on 15/Jun/20

p r o v e t h a t

\int_{0}^{\infty} \frac{3 + 2 \sqrt{x}}{x^{2} + 2 x + 5} d x = 4.13049

Commented by MJS last updated on 15/Jun/20

t = \sqrt{x} \to d x = 2 \sqrt{x} d t

2 \int \frac{t (2 t + 3)}{t^{4} + 2 t^{2} + 5} d t

now decompose (2 square factors) and solve .

as many times before, easy path but horrible

constants \dots

2 \int \frac{t (2 t + 3)}{t^{4} + 2 t^{2} + 5} d t =

= \frac{\sqrt{- 2 + 2 \sqrt{5}}}{4} (\int \frac{2 t + 3}{t^{2} - \sqrt{- 2 + 2 \sqrt{5}} t + \sqrt{5}} d t -

- \int \frac{2 t + 3}{t^{2} + \sqrt{- 2 + 2 \sqrt{5}} t + \sqrt{5}} d t) =

= \frac{\sqrt{- 2 + 2 \sqrt{5}}}{4} \times (

\int \frac{2 t - \sqrt{- 2 + 2 \sqrt{5}}}{t^{2} - \sqrt{- 2 + 2 \sqrt{5}} t + \sqrt{5}} d t +

+ (3 + \sqrt{- 2 + 2 \sqrt{5}}) \int \frac{d t}{t^{2} - \sqrt{- 2 + 2 \sqrt{5}} t + \sqrt{5}}) -

- \int \frac{2 t + \sqrt{- 2 + 2 \sqrt{5}}}{t^{2} + \sqrt{- 2 + 2 \sqrt{5}} t + \sqrt{5}} d t -

- (3 - \sqrt{- 2 + 2 \sqrt{5}}) \int \frac{d t}{t^{2} + \sqrt{- 2 + 2 \sqrt{5}} t + \sqrt{5}}

)

\dots and tbese are easy to solve

(I hope I made no error of sign in front of

the “ \int^{″} s \dots)

Commented by M±th+et+s last updated on 15/Jun/20

t h a n k y o u s i r

Answered by mathmax by abdo last updated on 15/Jun/20

I = \int_{0}^{\infty} \frac{3 + 2 \sqrt{x}}{x^{2} + 2 x + 5} dx changement \sqrt{x} = t give

I = \int_{0}^{\infty} \frac{3 + 2 t}{t^{4} + {2 t}^{2} + 5} (2 t) dt = 2 \int_{0}^{\infty} \frac{3 t + {2 t}^{2}}{t^{4} + {2 t}^{2} + 5} dt let decompose

F (t) = \frac{{2 t}^{2} + 3 t}{t^{4} + {2 t}^{2} + 5}

t^{4} + {2 t}^{2} + 5 = 0 \to u^{2} + 2 u + 5 = 0 \to Δ^{^{'}} = - 4 \Rightarrow u_{1} = - 1 + 2 i and u_{2} = - 1 - 2 i

\Rightarrow u^{2} + 2 u + 5 = (u - u_{1}) (u - u_{2}) = (t^{2} - u_{1}) (t^{2} - u_{2}) \Rightarrow

I = \int_{0}^{\infty} \frac{{2 t}^{2} + 3 t}{(t^{2} - u_{1}) (t^{2} - u_{2})} dt = \frac{1}{4 i} \int_{0}^{\infty} ({2 t}^{2} + 3 t) (\frac{1}{t^{2} - u_{1}} - \frac{1}{t^{2} - u_{2}}) dt

= \frac{1}{4 i} \int_{0}^{\infty} \frac{{2 t}^{2} + 3 t}{t^{2} - u_{1}} dt - \frac{1}{4 i} \int_{0}^{\infty} \frac{{2 t}^{2} + 3 t}{t^{2} - u_{2}} dt

= \frac{1}{4 i} \int_{0}^{\infty} \frac{2 (t^{2} - u_{1}) + 3 t + {2 u}_{1}}{t^{2} - u_{1}} dt - \frac{1}{4 i} \int_{0}^{\infty} \frac{2 (t^{2} - u_{2}) + 3 t + {2 u}_{2}}{t^{2} - u_{2}} dt

= \frac{1}{4 i} {\int_{0}^{\infty} \frac{3 t + {2 u}_{1}}{t^{2} - u_{1}} dt - \int_{0}^{\infty} \frac{3 t + {2 u}_{2}}{t^{2} - u_{2}} dt}

= \frac{1}{4 i} {\frac{3}{2} {[\ln (\frac{t^{2} - u_{1}}{t^{2} - u_{2}})]}_{0}^{+ \infty} + 2 \int_{0}^{\infty} \frac{u_{1}}{t^{2} - u_{1}} dt - 2 \int_{0}^{\infty} \frac{u_{2}}{t^{2} - u_{2}} dt}

= \frac{3}{8 i} \ln (\frac{u_{1}}{u_{2}}) + \frac{1}{2 i} {u_{1} \int_{0}^{\infty} \frac{dt}{t^{2} - u_{1}} - u_{2} \int_{0}^{\infty} \frac{dt}{t^{2} - u_{2}}} \dots . be continued \dots

Commented by M±th+et+s last updated on 15/Jun/20

t h a n k s s i r

Commented by mathmax by abdo last updated on 16/Jun/20

you are welcome .