find-dx-x-3-x-2-4x-

Question Number 66308 by mathmax by abdo last updated on 12/Aug/19

f i n d \int \frac{d x}{(x + 3) \sqrt{- x^{2} - 4 x}}

Commented by prof Abdo imad last updated on 16/Aug/19

l e t A = \int \frac{d x}{(x + 3) \sqrt{- x^{2} - 4 x}} w e h a v e

- x^{2} - 4 x = - (x^{2} + 4 x + 4 - 4) = 4 - {(x + 2)}^{2}

w e u s e t h e c h a n g e m e n t x + 2 = 2 s i n t \Rightarrow

A = \int \frac{2 c o s t d t}{(2 s i n t - 2 + 3) 2 c o s t} = \int \frac{d t}{2 s i n t + 1}

=_{t a n (\frac{t}{2}) = u} \int \frac{1}{(2 \frac{2 u}{1 + u^{2}} + 1)} \frac{2 d u}{1 + u^{2}}

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

u^{2} + 4 u + 1 = 0 \to Δ^{^{'}} = 4 - 1 = 3 \Rightarrow

u_{1} = - 2 + \sqrt{3} a n d u_{2} = - 2 - \sqrt{3} \Rightarrow

A = \int \frac{2 d u}{(u - u_{1}) (u - u_{2})} = \frac{1}{\sqrt{3}} \int (\frac{1}{u - u_{1}} - \frac{1}{u - u_{2}}) d u

= \frac{1}{\sqrt{3}} l n ∣ \frac{u - u_{1}}{u - u_{2}} ∣ + C = \frac{1}{\sqrt{3}} l n ∣ \frac{t a n (\frac{t}{2}) + 2 - \sqrt{3}}{t a n (\frac{t}{2}) + 2 + \sqrt{3}} ∣ + C

b u t t = a r c s i n (\frac{x + 2}{2}) \Rightarrow

A = \frac{1}{\sqrt{3}} l n ∣ \frac{t a n (\frac{1}{2} a r c s i n (\frac{x}{2} + 1) + 2 - \sqrt{3}}{t a n (\frac{1}{2} a r c s i n (\frac{x}{2} + 1)) + 2 + \sqrt{3}} ∣ + C