let-f-x-0-1-dt-1-x-1-t-2-with-x-gt-0-1-detemine-a-explicit-form-of-f-x-2-find-also-g-x-0-1-1-t-2-1-x-1-t-2-2-dt-3-find-the-value-of-integrals-0-1-dt-

Question Number 65773 by mathmax by abdo last updated on 03/Aug/19

l e t f (x) = \int_{0}^{1} \frac{d t}{1 + x \sqrt{1 + t^{2}}} w i t h x > 0

1) d e t e m i n e a e x p l i c i t f o r m o f f (x)

2) f i n d a l s o g (x) = \int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{{(1 + x \sqrt{1 + t^{2}})}^{2}} d t

3) f i n d t h e v a l u e o f i n t e g r a l s \int_{0}^{1} \frac{d t}{1 + 2 \sqrt{1 + t^{2}}} a n d

\int_{0}^{1} \frac{d t}{{(1 + 2 \sqrt{1 + t^{2}})}^{2}}

Commented by mathmax by abdo last updated on 06/Aug/19

2) w e h a v e f^{^{'}} (x) = - \int_{0}^{1} \frac{\sqrt{1 + t^{2}} d t}{{(1 + x \sqrt{1 + t^{2}})}^{2}} = - g (x) \Rightarrow g (x) = - f^{^{'}} (x)

r e s t t o c a l c u l a t e f^{^{'}} (x)

3) c h a n g e m e n t t = s h t g i v e I = \int_{0}^{1} \frac{d t}{1 + 2 \sqrt{1 + t^{2}}}

= \int_{0}^{l n (1 + \sqrt{2})} \frac{c h (t) d t}{1 + 2 c h (t)} = \int_{0}^{l n (1 + \sqrt{2})} \frac{\frac{e^{t} + e^{- t}}{2}}{1 + 2 \frac{e^{t} + e^{- t}}{2}} d t

= \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{t} + e^{- t}}{1 + e^{t} + e^{- t}} d t =_{e^{t} = x} \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{x + x^{- 1}}{1 + x + x^{- 1}} \frac{d x}{x}

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

= \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{x^{2} + 1}{x (x^{2} + x + 1)} d x l e t d e c o m p o s e F (x) = \frac{x^{2} + 1}{x (x^{2} + x + 1)}

F (x) = \frac{a}{x} + \frac{b x + c}{x^{2} + x + 1}

a = {l i m}_{x \to 0} x F (x) = 1

{l i m}_{x \to + \infty} x F (x) = 1 = a + b \Rightarrow b = 1 - a = 0 \Rightarrow F (x) = \frac{1}{x} + \frac{c}{x^{2} + x + 1}

F (1) = \frac{2}{3} = 1 + \frac{c}{3} \Rightarrow 2 = 3 + c \Rightarrow c = - 1 \Rightarrow F (x) = \frac{1}{x} - \frac{1}{x^{2} + x + 1}

\Rightarrow I = \frac{1}{2} {\int_{1}^{1 + \sqrt{2}} \frac{d x}{x} - \int_{1}^{1 + \sqrt{2}} \frac{d x}{x^{2} + x + 1}}

= \frac{1}{2} l n (1 + \sqrt{2}) - \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{d x}{x^{2} + x + 1}

\int_{1}^{1 + \sqrt{2}} \frac{d x}{x^{2} + x + 1} = \int_{1}^{1 + \sqrt{2}} \frac{d x}{{(x + \frac{1}{2})}^{2} + \frac{3}{4}} =_{x + \frac{1}{2} = \frac{\sqrt{3}}{2} u} \frac{4}{3} \int_{\sqrt{3}}^{\frac{3 + 2 \sqrt{2}}{3}} \frac{1}{1 + u^{2}} \frac{\sqrt{3}}{2} d u

= \frac{2}{\sqrt{3}} {[a r c t a n u]}_{\sqrt{3}}^{\frac{3 + 2 \sqrt{2}}{3}} = \frac{2}{\sqrt{3}} {a r c t a n (\frac{3 + 2 \sqrt{2}}{3}) - a r c t a n (\sqrt{3})} \Rightarrow

I = \frac{1}{2} l n (1 + \sqrt{2}) - \frac{1}{\sqrt{3}} {a r c t a n (\frac{3 + 2 \sqrt{2}}{3}) - a r c t a n (\sqrt{3})}

Commented by mathmax by abdo last updated on 06/Aug/19

1) f (x) = \int_{0}^{1} \frac{d t}{1 + x \sqrt{1 + t^{2}}} c h a n g e m e n t t = s h (u) g i v e

f (x) = \int_{0}^{l n (1 + \sqrt{2})} \frac{c h u}{1 + x c h u} d u = \int_{0}^{l n (1 + \sqrt{2})} \frac{\frac{e^{u} + e^{- u}}{2}}{1 + x \frac{e^{u} + e^{- u}}{2}} d u

= \int_{0}^{l n (1 + \sqrt{2})} \frac{e^{u} + e^{- u}}{2 + {x e}^{u} + x e^{- u}} d u =_{e^{u} = z} \int_{1}^{1 + \sqrt{2}} \frac{z + z^{- 1}}{2 + x z + {x z}^{- 1}} \frac{d z}{z}

= \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z^{2} (2 + x z + {x z}^{- 1})} d z = \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{2 z^{2} + {x z}^{3} + x z} d z

= \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z ({x z}^{2} + 2 z + x)} d z l e t d e c o m p o s e F (z) = \frac{z^{2} + 1}{z ({x z}^{2} + 2 z + x)}

{x z}^{2} + 2 z + x = 0 \to Δ^{^{'}} = 1 - x^{2}

c a s e 1 1 - x^{2} > 0 \Rightarrow 0 < x < 1 \Rightarrow z_{1} = \frac{- 1 + \sqrt{1 - x^{2}}}{x}

z_{2} = \frac{- 1 - \sqrt{1 - x^{2}}}{x} \Rightarrow F (z) = \frac{z^{2} + 1}{x z (z - z_{1}) (z - z_{2})}

= \frac{a}{z} + \frac{b}{z - z_{1}} + \frac{c}{z - z_{2}}

a = {l i m}_{z \to 0} z F (z) = \frac{1}{x}

b = {l i m}_{z \to z_{1}} (z - z_{1}) F (z) = \frac{z_{1}^{2} + 1}{{x z}_{1} (z_{1} - z_{2})}

c = {l i m}_{z \to z_{2}} (z - z_{2}) F (z) = \frac{z_{2}^{2} + 1}{{x z}_{2} (z_{2} - z_{1})} \Rightarrow

f (x) = \int_{1}^{1 + \sqrt{2}} F (z) d z = {[a l n ∣ z ∣ + b l n ∣ z - z_{1} ∣ + c l n ∣ z - z_{2} ∣]}_{1}^{1 + \sqrt{2}}

= a l n (1 + \sqrt{2}) + b l n ∣ 1 + \sqrt{2} - z_{1} ∣ + c l n ∣ 1 + \sqrt{2} - z_{2} ∣ - b l n ∣ 1 - z_{1} ∣

- c l n ∣ 1 - z_{2} ∣

c a s e 2 1 - x^{2} < 0 \Rightarrow x > 1 f (x) = \int_{1}^{1 + \sqrt{2}} \frac{z^{2} + 1}{z ({x z}^{2} + 2 z + x)} d z

F (z) = \frac{a}{z} + \frac{b z + c}{{x z}^{2} + 2 z + x}

a = \frac{1}{x}

{l i m}_{z \to + \infty} z F (z) = \frac{1}{x} = a + \frac{b}{x} \Rightarrow 1 = a x + b \Rightarrow b = 1 - a x = 0 \Rightarrow

F (z) = \frac{1}{x z} + \frac{c}{{x z}^{2} + 2 z + x}

F (1) = \frac{2}{(2 x + 2)} = \frac{1}{x + 1} = \frac{1}{x} + \frac{c}{2 x + 2} \Rightarrow 1 = \frac{x + 1}{x} + \frac{c}{2} \Rightarrow

\frac{c}{2} = - \frac{1}{x} \Rightarrow c = \frac{- 2}{x} \Rightarrow F (z) = \frac{1}{x z} - \frac{2}{x} \frac{1}{({x z}^{2} + 2 z + x)}

= \frac{1}{x} {\frac{1}{z} - \frac{2}{{x z}^{2} + 2 z + x}} \Rightarrow f (x) = \frac{1}{x} {\int_{1}^{1 + \sqrt{2}} \frac{d z}{z} - 2 \int_{1}^{1 + \sqrt{2}} \frac{d z}{{x z}^{2} + 2 z + x}}

= \frac{1}{x} l n (1 + \sqrt{2}) - \frac{2}{x} \int_{1}^{1 + \sqrt{2}} \frac{d z}{{x z}^{2} + 2 z + x} = \dots .