let-f-x-0-1-dt-t-x-t-2-1-x-real-parametre-1-find-a-explicite-form-forf-x-2-detemine-also-g-x-0-1-dt-t-x-t-2-1-2-3-give-f-n-x-at-form-of-integrals-4-

Question Number 64429 by mathmax by abdo last updated on 17/Jul/19

l e t f (x) = \int_{0}^{1} \frac{d t}{t + x + \sqrt{t^{2} + 1}} (x r e a l p a r a m e t r e)

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

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

3) g i v e f^{(n)} (x) a t f o r m o f i n t e g r a l s

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

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

Commented by mathmax by abdo last updated on 20/Jul/19

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

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

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

=_{e^{u} = z} \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{z + z^{- 1}}{x + z} \frac{d z}{z} = \frac{1}{2} \int_{1}^{1 + \sqrt{2}} \frac{z + z^{- 1}}{x z + z^{2}} d z

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

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

b = {l i m}_{z \to 0} z^{2} F (z) = \frac{1}{x}

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

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

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

2 = (x + 1) a + \frac{x^{2} + x + x^{2} + 1}{x^{2}} = (x + 1) a + \frac{2 x^{2} + x + 1}{x^{2}} \Rightarrow

2 x^{2} = x^{2} (x + 1) a + 2 x^{2} + x + 1 \Rightarrow x^{2} (x + 1) a = - (x + 1) \Rightarrow

a = - \frac{1}{x^{2}} (i f x \neq - 1) \Rightarrow F (z) = - \frac{1}{x^{2} z} + \frac{1}{{x z}^{2}} + \frac{x^{2} + 1}{x^{2} (z + x)} \Rightarrow

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

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

= - \frac{1}{x^{2}} {[l n ∣ z ∣]}_{1}^{1 + \sqrt{2}} - \frac{1}{x} {[\frac{1}{z}]}_{1}^{1 + \sqrt{2}} + \frac{x^{2} + 1}{x^{2}} {[l n ∣ z + x ∣]}_{1}^{1 + \sqrt{2}}

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

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

Commented by mathmax by abdo last updated on 20/Jul/19

2) t h e o r e m o f d e r i v a t i o n g i v e

f^{^{'}} (x) = \int_{0}^{1} \frac{\partial}{\partial x} (\frac{1}{t + x + \sqrt{t^{2} + 1}}) d t = - \int_{0}^{1} \frac{d t}{{(t + x + \sqrt{t^{2} + 1})}^{2}} = - g (x) \Rightarrow

g (x) = - f^{^{'}} (x) f i s k n o w n r e s t t o c a l c u l a t e f^{^{'}} (x)

Commented by mathmax by abdo last updated on 20/Jul/19

3) w e h a v e f^{(n)} (x) = \int_{0}^{1} \frac{\partial^{n}}{\partial x^{n}} (\frac{1}{t + x + \sqrt{t^{2} + 1}}) d t

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

Commented by mathmax by abdo last updated on 20/Jul/19

4) \int_{0}^{1} \frac{d t}{t + \sqrt{t^{2} + 1}} =_{t = s h (u)} \int_{0}^{l n (1 + \sqrt{2})} \frac{c h u}{s h (u) + c h (u)} d u

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

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

= \frac{l n (1 + \sqrt{2})}{2} - \frac{1}{4} {[e^{- 2 u}]}_{0}^{l n (1 + \sqrt{2})}

= \frac{l n (1 + \sqrt{2})}{2} - \frac{1}{4} {\frac{1}{{(1 + \sqrt{2})}^{2}} - 1} .

Commented by mathmax by abdo last updated on 20/Jul/19

5) \int_{0}^{1} \frac{d t}{t + 1 + \sqrt{t^{2} + 1}} = f (1) = \frac{1}{2} {- l n (1 + \sqrt{2}) + \sqrt{2} + 2 l n ∣ \frac{2 + \sqrt{2}}{2} ∣