fnd-dx-x-2-3-x-2- – Tinku Tara

Question Number 65920 by mathmax by abdo last updated on 05/Aug/19

f n d \int \frac{d x}{x + 2 - \sqrt{3 + x^{2}}}

Commented by mathmax by abdo last updated on 12/Aug/19

l e t I = \int \frac{d x}{x + 2 - \sqrt{3 + x^{2}}} c h a 7 g e m e n t x = \sqrt{3} s h (t) g i v e

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

= \int \frac{\sqrt{3} (e^{t} + e^{- t})}{\sqrt{3} (e^{t} - e^{- t}) + 4 - \sqrt{3} (e^{t} + e^{- t})} d t

=_{e^{t} = u} \int \frac{\sqrt{3} (u + u^{- 1})}{\sqrt{3} (u - u^{- 1}) + 4 - \sqrt{3} (u + u^{- 1})} \frac{d u}{u}

= \int \frac{\sqrt{3} (u + u^{- 1})}{\sqrt{3} u^{2} - \sqrt{3} + 4 u - \sqrt{3} u^{2} - \sqrt{3}} d u = \int \frac{\sqrt{3} (u + u^{- 1})}{4 u - 2 \sqrt{3}} d u

= \int \frac{\sqrt{3} u^{2} + \sqrt{3}}{4 u^{2} - 2 \sqrt{3} u} d u = \frac{\sqrt{3}}{4} \int \frac{u^{2} + 1}{u^{2} - \frac{\sqrt{3}}{2}} d u

= \frac{\sqrt{3}}{4} \int \frac{u^{2} - \frac{\sqrt{3}}{2} + 1 - \frac{\sqrt{3}}{2}}{u^{2} - \frac{\sqrt{3}}{2}} d u = \frac{\sqrt{3}}{4} u + \frac{\sqrt{3}}{8} (2 - \sqrt{3}) \int \frac{d u}{u^{2} - \frac{\sqrt{3}}{2}}

= \frac{\sqrt{3}}{4} u + \frac{\sqrt{3}}{8} (2 - \sqrt{3}) \int (\frac{1}{u - \sqrt{\frac{\sqrt{3}}{2}}} - \frac{1}{u + \sqrt{\frac{\sqrt{3}}{2}}}) d u

= \frac{\sqrt{3}}{4} u + \frac{\sqrt{3}}{8} (2 - \sqrt{3}) l n ∣ \frac{u - \sqrt{\frac{\sqrt{3}}{2}}}{u + \sqrt{\frac{\sqrt{3}}{2}}} ∣ + c w e h a v e

u = e^{t} a n d t = a r g s h (\frac{x}{\sqrt{3}}) = l n (\frac{x}{\sqrt{3}} + \sqrt{1 + \frac{x^{2}}{3}}) \Rightarrow u = \frac{x + \sqrt{3 + x^{2}}}{\sqrt{3}}

\Rightarrow I = \frac{\sqrt{3} (x + \sqrt{3 + x^{2}})}{4 \sqrt{3}} + \frac{\sqrt{3}}{8} (2 - \sqrt{3}) l n ∣ \frac{\frac{x + \sqrt{3 + x^{2}}}{\sqrt{3}} - \sqrt{\frac{\sqrt{3}}{2}}}{\frac{x + \sqrt{3 + x^{2}}}{\sqrt{3}} + \sqrt{\frac{\sqrt{3}}{2}}} ∣ + C

Answered by MJS last updated on 07/Aug/19

\int \frac{d x}{x + 2 - \sqrt{x^{2} + 3}} = 2 \int \frac{d x}{4 x + 1} + \int \frac{x}{4 x + 1} d x + 3 \int \frac{\sqrt{x^{2} + 3}}{4 x + 1} d x

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

\int \frac{x}{4 x + 1} d x = \frac{1}{4} \int d x - \frac{1}{4} \int \frac{d x}{4 x + 1} = \frac{x}{4} - \frac{1}{16} \ln (4 x + 1)

3 \int \frac{\sqrt{x^{2} + 3}}{4 x + 1} d x =

[t = \arctan (\frac{\sqrt{3}}{3} x) \to d x = \frac{\sqrt{3}}{3} (x^{2} + 3) d t]

= 9 \int \frac{d t}{(4 \sqrt{3} \sin t + \cos t) \cos^{2} t} =

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

= - 18 \int \frac{{(u^{2} + 1)}^{2}}{{(u - 1)}^{2} {(u + 1)}^{2} (u - 7 - 4 \sqrt{3}) (u + 7 - 4 \sqrt{3})} d u =

= \frac{3 (1 - 4 \sqrt{3})}{16} \int \frac{d u}{u - 1} + \frac{3 \sqrt{3}}{4} \int \frac{u}{{(u - 1)}^{2}} d u - \frac{3 (1 + 4 \sqrt{3})}{16} \int \frac{d u}{u + 1} + \frac{3 \sqrt{3}}{4} \int \frac{u}{{(u + 1)}^{2}} d u - \frac{21}{16} \int \frac{d u}{u - 7 - 4 \sqrt{3}} + \frac{21}{16} \int \frac{d u}{u + 7 - 4 \sqrt{3}} =

now {it}^{'} s easy \dots

= \frac{21}{16} \ln \frac{u + 7 - 4 \sqrt{3}}{u - 7 - 4 \sqrt{3}} + \frac{3}{16} \ln \frac{u - 1}{u + 1} - \frac{3 \sqrt{3}}{2 u^{2} - 2} =

[u = \tan \frac{t}{2} = \frac{\sqrt{x^{2} + 3} - \sqrt{3}}{x}]

= \frac{21}{16} \ln ∣ \frac{x - 12 + 7 \sqrt{x^{2} + 3}}{4 x + 1} ∣ + \frac{3}{16} \ln ∣ x - \sqrt{x^{2} + 3} ∣ + \frac{3}{4} \sqrt{x^{2} + 3}

\Rightarrow

\int \frac{d x}{x + 2 - \sqrt{x^{2} + 3}} = \frac{7}{16} \ln ∣ 4 x + 1 ∣ + \frac{21}{16} \ln ∣ \frac{x - 12 + 7 \sqrt{x^{2} + 3}}{4 x + 1} ∣ + \frac{3}{16} \ln ∣ x - \sqrt{x^{2} + 3} ∣ + \frac{1}{4} (x + 3 \sqrt{x^{2} + 3}) + C

\dots I hope I made no typos \dots

Commented by mathmax by abdo last updated on 07/Aug/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 d b u t y o u c a n u s e t h e

c h a n g e m e n t x = \sqrt{3} s h (t) \dots i t s e a z y i n t h i s c a s e \dots

Commented by MJS last updated on 07/Aug/19

true \dots

but it was great fun to solve it like this \dots

Commented by MJS last updated on 07/Aug/19

I then get

\frac{7}{8} \ln ∣ 2 x - 3 + 2 \sqrt{x^{2} + 3} ∣ - \frac{1}{2} arcsinh (\frac{\sqrt{3}}{3} x) + \frac{1}{4} (x + \sqrt{x^{2} + 3}) + C

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