Question-214072

Question Number 214072 by Mwamba last updated on 26/Nov/24

Answered by TonyCWX08 last updated on 27/Nov/24

Answered by TonyCWX08 last updated on 27/Nov/24

C o n s i d e r t w o p o i n t s, P a n d Q .

P = (x, y) a n d Q = (x + δ x, y + δ y)

L e t s = l e n g t h o f a r c f r o m s t a r t i n g p o i n t

δ s = L e n g t h o f a r c P Q

S i n c e δ s i s v e r y s m a l l, w e c a n a p p r o x i m a t e a r c P Q t o a s t r a i g h t l i n e .

B y {P y t h a g o r a s}^{'} s t h e o r e m,

{(δ x)}^{2} + {(δ y)}^{2} = {(δ s)}^{2}

D i v i d i n g b o t h s i d e s b y {(δ x)}^{2}, w e g e t

1 + {(\frac{δ y}{δ x})}^{2} = {(\frac{δ s}{δ x})}^{2}

A s δ x \to 0, t h i s g i v e s

1 + {(\frac{d y}{d x})}^{2} = {(\frac{d s}{d x})}^{2}

A p p l y i n g S q u a r e R o o t o n b o t h s i d e s,

\frac{d s}{d x} = \sqrt{1 + {(\frac{d y}{d x})}^{2}}

T h e r e f o r e, w e h a v e

s = \int \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x o r \int \sqrt{1 + {(f^{'} (x))}^{2}} d x

Answered by TonyCWX08 last updated on 27/Nov/24

\int \frac{d x}{x^{4} + 5 x^{2} + 6}

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

= \int (- \frac{1}{x^{2} + 3} + \frac{1}{x^{2} + 2}) d x

= - \int (\frac{1}{x^{2} + 3}) d x + \int (\frac{1}{x^{2} + 2}) d x

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

Answered by lepuissantcedricjunior last updated on 28/Nov/24

\boldsymbol b) \frac{\boldsymbol π (3 \sqrt{2} - 2 \sqrt{3})}{12}

b) \frac{π (3 \sqrt{2} - 2 \sqrt{3})}{12}