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

Question Number 58220 by maxmathsup by imad last updated on 20/Apr/19

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

Answered by tanmay last updated on 20/Apr/19

- x^{2} - x + 3 x + 3

- x (x + 1) + 3 (x + 1)

(x + 1) (3 - x)

t^{2} = x + 1 d x = 2 t d t

\int \frac{d x}{x (x + 1) \sqrt{(x + 1) (3 - x)}}

\int \frac{2 t d t}{(t^{2} - 1) t^{2} \sqrt{t^{2} (3 - t^{2} + 1)}}

\int \frac{2 d t}{(t^{2} - 1) t^{2} \sqrt{4 - t^{2}}}

2 \int \frac{t^{2} - (t^{2} - 1)}{t^{2} (t^{2} - 1) \sqrt{4 - t^{2}}} d t

2 \int \frac{d t}{(t^{2} - 1) \sqrt{4 - t^{2}}} - 2 \int \frac{d t}{t^{2} \sqrt{4 - t^{2}}}

2 I_{1} - 2 I_{2}

I_{1} = \int \frac{d t}{(t^{2} - 1) \sqrt{4 - t^{2}}}

a^{2} = \frac{4 - t^{2}}{t^{2} - 1} \to a^{2} t^{2} - a^{2} = 4 - t^{2}

t^{2} (a^{2} + 1) = a^{2} + 4

t^{2} = \frac{a^{2} + 4}{a^{2} + 1}

2 t d t = \frac{(a^{2} + 1) (2 a) - (a^{2} + 4) (2 a)}{{(a^{2} + 1)}^{2}} d a

2 t d t = \frac{2 a^{3} + 2 a - 2 a^{3} - 8 a}{{(a^{2} + 1)}^{2}} d a

t d t = \frac{- 3 a d a}{{(a^{2} + 1)}^{2}}

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

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

\int \frac{- 3 a d a}{\sqrt{a^{2} + 4}} \times \frac{1}{3 \sqrt{3} a}

\frac{- 1}{\sqrt{3}} \int \frac{d a}{\sqrt{a^{2} + 4}} \to {\frac{- 1}{\sqrt{3}} l n (a + \sqrt{a^{2} + 4})}

s o 2 I_{1} = \frac{- 2}{\sqrt{3}} l n (\sqrt{\frac{4 - t^{2}}{t^{2} - 1}} + \sqrt{\frac{4 - t^{2}}{t^{2} - 1} + 4})

= \frac{- 2}{\sqrt{3}} l n (\frac{\sqrt{4 - t^{2}} + t \sqrt{3}}{\sqrt{t^{2} - 1}}) \leftarrow v a l u e o f I_{1}

l e n g t h y p r o b l e m w a i t f o r I_{2}

I_{2} = \int \frac{d t}{t^{2} \sqrt{4 - t^{2}}}

t = 2 s i n θ

\int \frac{2 c o s θ d θ}{4 {s i n}^{2} θ \times 2 c o s θ}

= \frac{1}{2} \int {c o s e c}^{2} θ d θ

= \frac{- c o t θ}{2} = \frac{- 1}{2} (\frac{\sqrt{4 - t^{2}}}{t}) \leftarrow v a l u e o f I_{2}

s o c o m p l e t e a n s w d r i s 2 I_{1} - 2 I_{2}

2 \times \frac{- 2}{\sqrt{3}} l n (\frac{\sqrt{4 - t^{2}} + t \sqrt{3}}{\sqrt{t^{2} - 1}}) - 2 \times \frac{- 1}{2} (\frac{\sqrt{4 - t^{2}}}{t}) + c

t^{2} = x + 1

= \frac{- 4}{\sqrt{3}} l n (\frac{\sqrt{3 - x} + \sqrt{3 (x + 1)}}{\sqrt{x}}) + (\frac{\sqrt{3 - x}}{\sqrt{x + 1}}) + c

p l s c h e c k \dots

Facebook Tweet Pin