find-f-dx-x-x-2-3-and-g-dx-x-x-2-3-2-with-real-

Question Number 69792 by mathmax by abdo last updated on 27/Sep/19

f i n d f (α) = \int \frac{d x}{x + α + \sqrt{x^{2} + 3}}

a n d g (α) = \int \frac{d x}{{(x + α + \sqrt{x^{2} + 3})}^{2}} w i t h α r e a l

Commented by mathmax by abdo last updated on 10/Oct/19

f (α) = \int \frac{d x}{x + α + \sqrt{x^{2} + 3}} c h a n g e m e n t x = \sqrt{3} s h (t) g i v e

f (α) = \int \frac{\sqrt{3} c h (t) d t}{\sqrt{3} s h (t) + α + \sqrt{3} c h (t)} = \int \frac{\frac{e^{t} + e^{- t}}{2}}{\frac{e^{t} - e^{- t}}{2} + α + \frac{e^{t} + e^{- t}}{2}} d t

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

=_{e^{t} = u} \frac{1}{2} \int \frac{u + u^{- 1}}{u + α} \frac{d u}{u} = \frac{1}{2} \int \frac{u^{2} + 1}{u^{2} (u + α)} d u l e t d e c o m p o s e

F (u) = \frac{u^{2} + 1}{u^{2} (u + α)} \Rightarrow F (u) = \frac{a}{u} + \frac{b}{u + α} + \frac{c}{u^{2}}

b = {l i m}_{u \to - α} (u + α) F (u) = \frac{1 + α^{2}}{α^{2}}

c = {l i m}_{u \to 0} u^{2} F (u) = \frac{1}{α} \Rightarrow F (u) = \frac{a}{u} + \frac{1 + α^{2}}{α^{2} (u + α)} + \frac{1}{α u^{2}}

{l i m}_{u \to + \infty} u F (u) = 1 = a + \frac{1 + α^{2}}{α^{2}} \Rightarrow a = 1 - \frac{1}{α^{2}} - 1 = - \frac{1}{α^{2}} \Rightarrow

F (u) = - \frac{1}{α^{2} u} + \frac{1 + α^{2}}{α^{2} (u + α)} + \frac{1}{α u^{2}} \Rightarrow

\int F (u) d u = - \frac{1}{α^{2}} l n ∣ u ∣ + \frac{1 + α^{2}}{α^{2}} l n ∣ u + α ∣ - \frac{1}{α u} + c

= - \frac{t}{α^{2}} + \frac{1 + α^{2}}{α^{2}} l n ∣ e^{t} + α ∣ - \frac{1}{α} e^{- t} + c b u t t = a r g s h (\frac{x}{\sqrt{3}})

= l n (\frac{x}{\sqrt{3}} + \sqrt{1 + \frac{x^{2}}{3}}) \Rightarrow

\int F (u) d u = - \frac{1}{α^{2}} l n (\frac{x}{\sqrt{3}} + \sqrt{1 + \frac{x^{2}}{3}}) + \frac{1 + α^{2}}{α^{2}} l n ∣ \frac{x}{3} + \sqrt{1 + \frac{x^{2}}{3}} + α ∣

- \frac{1}{α (\frac{x}{3} + \sqrt{1 + \frac{x^{2}}{3}})} + C \Rightarrow

f (α) = - \frac{1}{2 α^{2}} l n (\frac{x}{\sqrt{3}} + \sqrt{1 + \frac{x^{2}}{3}}) + \frac{1 + α^{2}}{2 α^{2}} l n ∣ \frac{x}{3} + \sqrt{1 + \frac{x^{2}}{3}} + α ∣

- \frac{1}{2 α (\frac{x}{3} + \sqrt{1 + \frac{x^{2}}{3}})} + C .

Answered by MJS last updated on 28/Sep/19

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

[we have to substitute x = \sqrt{3} \sinh t and

then t = \ln u

\Rightarrow both in 1 step gives

u = \frac{\sqrt{3}}{3} (x + \sqrt{x^{2} + 3}) \to d x = \frac{\sqrt{3} (u^{2} + 1)}{2 u^{2}} d u

x = \frac{\sqrt{3} (u^{2} - 1)}{2 u}]

= \frac{\sqrt{3}}{2} \int \frac{u^{2} + 1}{u^{2} (\sqrt{3} u + α)} d u =

= \frac{\sqrt{3}}{2 α} \int \frac{d u}{u^{2}} - \frac{3}{2 α^{2}} \int \frac{d u}{u} + \frac{\sqrt{3} (α^{2} + 3)}{2 α^{2}} \int \frac{d u}{\sqrt{3} u + α} =

= - \frac{\sqrt{3}}{2 α} u^{- 1} - \frac{3}{2 α^{2}} \ln u + \frac{α^{2} + 3}{2 α^{2}} \ln (\sqrt{3} u + α) =

= \frac{1}{2 α} (x - \sqrt{x^{2} + 3}) - \frac{3}{2 α^{2}} \ln (x + \sqrt{x^{2} + 3}) + \frac{α^{2} + 3}{2 α^{2}} \ln (x + α + \sqrt{x^{2} + 3}) + C

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

[same substitution as above]

= \frac{\sqrt{3}}{2} \int \frac{u^{2} + 1}{u^{2} {(\sqrt{3} u + α)}^{2}} d u =

= \frac{\sqrt{3}}{2 α^{2}} \int \frac{d u}{u^{2}} - \frac{3}{α^{3}} \int \frac{d u}{u} + \frac{\sqrt{3} (α^{2} + 3)}{2 α^{2}} \int \frac{d u}{{(\sqrt{3} u + α)}^{2}} + \frac{3 \sqrt{3}}{α^{3}} \int \frac{d u}{\sqrt{3} u + α} =

= - \frac{\sqrt{3}}{2 α^{2}} u^{- 1} - \frac{3}{α^{3}} \ln u - \frac{α^{2} + 3}{2 α^{2} (\sqrt{3} u + α)} + \frac{3}{α^{3}} \ln (\sqrt{3} u + α) =

= \frac{1}{2 α^{2}} (x - \sqrt{x^{2} + 3}) - \frac{3}{α^{3}} \ln (x + \sqrt{x^{2} + 3}) - \frac{α^{2} + 3}{2 α^{2} (x + α + \sqrt{x^{2} + 3})} + \frac{3}{α^{3}} \ln (x + α + \sqrt{x^{2} + 3}) + C =

= \frac{2 α x^{2} - 6 x - α (α^{2} + 3) - 2 (α x - 3) \sqrt{x^{2} + 3}}{2 α^{2} (2 α x + α^{2} - 3)} + \frac{3}{α^{3}} \ln (- α x + 3 + α \sqrt{x^{2} + 3}) + C

Commented by mathmax by abdo last updated on 28/Sep/19

t h a n k y o u s i r m j s f o r t h i s h a r d w o r k .