calculate-0-dx-x-1-2-x-2-4-

Question Number 71665 by mathmax by abdo last updated on 18/Oct/19

c a l c u l a t e \int_{0}^{\infty} \frac{d x}{{(x + 1)}^{2} (\sqrt{x^{2} + 4})}

Answered by MJS last updated on 19/Oct/19

\int \frac{d x}{{(x + 1)}^{2} \sqrt{x^{2} + 4}} =

[t = \arctan \frac{x}{2} \to d x = \frac{x^{2} + 4}{2} d t]

= \int \frac{\cos t}{{(\cos t + 2 \sin t)}^{2}} d t =

[u = \tan \frac{t}{2} \to d t = \frac{2 d u}{u^{2} + 1}]

= - 2 \int \frac{u^{2} - 1}{{(u^{2} - 4 u - 1)}^{2}}

now I^{'} d use {Ostrogradski}^{'} s Method once again

\int \frac{P}{Q} = \frac{P_{1}}{Q_{1}} + \int \frac{P_{2}}{Q_{2}}

Q_{1} = \gcd (Q, Q^{'}); Q_{2} = \frac{Q}{Q_{1}}

degree (P_{i}) < degree (Q_{i})

we find P_{i} by comparing the factors of

\frac{P}{Q} = {(\frac{P_{1}}{Q_{1}})}^{'} + \frac{P_{2}}{Q_{2}}

Q = {(u^{2} - 4 u - 1)}^{2}

Q^{'} = 4 (u - 2) (u^{2} - 4 u - 1)

Q_{1} = Q_{2} = u^{2} - 4 u - 1

P = u^{2} - 1

P_{1} = - \frac{4}{5} u - \frac{2}{5}

P_{2} = \frac{1}{5}

- 2 \int \frac{u^{2} - 1}{{(u^{2} - 4 u - 1)}^{2}} =

= \frac{4 (2 u + 1)}{5 (u^{2} - 4 u - 1)} - \frac{2}{5} \int \frac{d u}{u^{2} - 4 u - 1} =

= \frac{4 (2 u + 1)}{5 (u^{2} - 4 u - 1)} - \frac{\sqrt{5}}{25} \ln \frac{u - 2 - \sqrt{5}}{u - 2 + \sqrt{5}} =

= \dots = - \frac{\sqrt{x^{2} + 4}}{5 (x + 1)} + \frac{\sqrt{5}}{25} \ln ∣ \frac{x - 4 - \sqrt{5 (x^{2} + 4)}}{x + 1} ∣ + C

\underset{0}{\int^{\infty}} \frac{d x}{{(x + 1)}^{2} \sqrt{x^{2} + 4}} = \frac{1}{5} + \frac{\sqrt{5}}{25} \ln (\frac{7 - 3 \sqrt{5}}{2})

Commented by Abdo msup. last updated on 19/Oct/19

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