find-f-a-1-2-arctan-x-a-x-dx-and-calculate-f-a-at-form-of-integral-

Question Number 68040 by mathmax by abdo last updated on 03/Sep/19

f i n d f (a) = \int_{1}^{2} a r c t a n (x + \frac{a}{x}) d x a n d

c a l c u l a t e f^{^{'}} (a) a t f o r m o f i n t e g r a l

Commented by mathmax by abdo last updated on 04/Sep/19

f (a) = \int_{1}^{2} a r c t a n (x + \frac{a}{x}) d x b y p a r t s

f (a) = {[x a r c t a n (x + \frac{a}{x})]}_{1}^{2} - \int_{1}^{2} x \frac{1 - \frac{a}{x^{2}}}{1 + {(x + \frac{a}{x})}^{2}} d x

= 2 a r c t a n (2 + \frac{a}{2}) - a r c t a n (1 + a) - \int_{1}^{2} \frac{x (x^{2} - a)}{x^{2} + {(x^{2} + a)}^{2}} d x

l e t J = \int_{1}^{2} \frac{x^{3} - a x}{x^{2} + {(x^{2} + a)}^{2}} d x \Rightarrow J = \int_{1}^{2} \frac{x^{3} - a x}{x^{2} + x^{4} + 2 {a x}^{2} + a^{2}} d x

= \int_{1}^{2} \frac{x^{3} - a x}{x^{4} + (2 a + 1) x^{2} + a^{2}}

x^{4} + (2 a + 1) x^{2} + a^{2} = 0 \Rightarrow t^{2} + (2 a + 1) t + a^{2} = 0 w i t h t = x^{2}

Δ = {(2 a + 1)}^{2} - 4 a^{2} = 4 a^{2} + 4 a + 1 - 4 a^{2} = 4 a + 1

c a s e 1 i f 4 a + 1 ⩾ 0 \Rightarrow t_{1} = \frac{- (2 a + 1) + \sqrt{4 a + 1}}{2}

a n d t_{2} = \frac{- (2 a + 1) - \sqrt{4 a + 1}}{2} \Rightarrow x^{4} + (2 a + 1) x^{2} + a^{2}

= (t - t_{1}) (t - t_{2}) (x^{2} - t_{1}) (x^{2} - t_{2}) \Rightarrow

J = \frac{1}{t_{1} - t_{2}} \int_{1}^{2} (x^{3} - a x) {\frac{1}{x^{2} - t_{1}} - \frac{1}{x^{2} - t_{2}}} d x

= \frac{1}{\sqrt{4 a + 1}} {\int_{1}^{2} \frac{x^{3} - a x}{x^{2} - t_{1}} d x - \int_{1}^{2} \frac{x^{3} - a x}{x^{2} - t_{2}} d x} w e h a v e

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

= {[\frac{x^{2}}{2}]}_{1}^{2} + (t_{1} - a) \int_{1}^{2} \frac{x d x}{x^{2} - t_{1}} = 2 - \frac{1}{2} + \frac{t_{1} - a}{2} {[l n ∣ x^{2} - t_{1} ∣]}_{1}^{2}

= \frac{3}{2} + \frac{t_{1} - a}{2} {l n ∣ \frac{4 - t_{1}}{1 - t_{1}} ∣} a l s o w e g e t

\int_{1}^{2} \frac{x^{3} - a x}{x^{2} - t_{2}} d x = \frac{3}{2} + \frac{t_{2} - a}{2} l n ∣ \frac{4 - t_{2}}{1 - t_{2}} ∣ \Rightarrow

J = \frac{1}{\sqrt{4 a + 1}} {\frac{(t_{1} - a)}{2} l n ∣ \frac{4 - t_{1}}{1 - t_{1}} ∣ - \frac{t_{2} - a}{2} l n ∣ \frac{4 - t_{2}}{1 - t_{2}} ∣} \Rightarrow

f (a) = 2 a r c t a n (2 + \frac{a}{2}) - a r c t a n (1 + a)

- \frac{1}{\sqrt{4 a + 1}} {\frac{t_{1} - a}{2} l n ∣ \frac{4 - t_{1}}{1 - t_{1}} ∣ - \frac{t_{2} - a}{2} l n ∣ \frac{4 - t_{2}}{1 - t_{2}} ∣} .

r e s t t o c a l c u l a t e f (a) i f 4 a + 1 < 0 ? ? \dots . b e c o n t i n u e d \dots

Commented by mathmax by abdo last updated on 05/Sep/19

f^{^{'}} (a) = \int_{1}^{2} \frac{d x}{x (1 + {(x + \frac{a}{x})}^{2})} = \int_{1}^{2} \frac{x^{2} d x}{x {x^{2} + {(x^{2} + a)}^{2}}}

= \int_{1}^{2} \frac{x d x}{x^{2} + x^{4} + 2 {a x}^{2} + a^{2}} = \int_{1}^{2} \frac{x d x}{x^{4} + (2 a + 1) x^{2} + a^{2}}

Answered by mind is power last updated on 03/Sep/19

i n t e g r a t i o n b y p a r t

f (s p a) = [x a r c t g (x + \frac{a}{x})] - \int x \frac{(1 - \frac{a}{x^{2}})}{1 + {(x + \frac{a}{x})}^{2}} d x

= 2 a r c t g (2 + \frac{a}{2}) - a r c t g (1 + a) - \int_{1}^{2} \frac{x (x^{2} - a)}{x^{4} + (2 a + 1) x^{2} + a^{2}} d x

u = x^{2}

d u = 2 x d x

f (a) = 2 a r c t v (2 + \frac{a}{2}) - a r c t g (1 + a) - \int_{1}^{4} \frac{(u - a) d u}{2 (u^{2} + (2 a + 1) u + a^{2})}

= 2 a r c t g (2 + \frac{a}{2}) - a r c t g (1 + a) - \int_{1}^{4} \frac{u + (a + \frac{1}{2}) - 2 a - \frac{1}{2}}{2 (u^{2} + (2 a + 1) u + a^{2}} d u

= 2 a r c t g (2 + \frac{a}{2}) - a r c t g (1 + a) - \int_{1}^{4} \frac{u + (a + \frac{1}{2})}{2 (u^{2} + (2 a + 1) u + a^{2}} d u + \frac{4 a + 1}{2} \int_{1}^{4} \frac{1}{{(u + \frac{2 a + 1}{2})}^{2} + a + \frac{1}{4}} d u

a > - \frac{1}{4}

= 2 a r c t g (2 + \frac{a}{2}) - a r c t g (1 + a) - \frac{1}{4} {[l n (u^{2} + (2 a + 1) u + a^{2})]}_{1}^{4} + 2 \int_{1}^{4} \frac{d u}{{(\frac{2 u}{\sqrt{4 a + 1}} + \frac{2 a + 1}{\sqrt{4 a + 1}})}^{2} + 1}

= 2 a r c t g (2 + \frac{a}{2}) - a r c t g (1 + a) - \frac{1}{4} l n (\frac{a^{2} + 8 a + 20}{a^{2} + 2 a + 2}) + \sqrt{4 a + 1} {[a r c t g (\frac{2 u}{\sqrt{4 a + 1}} + \frac{2 a + 1}{\sqrt{4 a + 1}})]}_{1}^{2}

b e e c o n t i n u e d

Commented by mathmax by abdo last updated on 04/Sep/19

t h a n k s s i r .