let-f-x-0-1-ln-1-xt-2-t-2-dt-with-x-R-1-find-a-explicit-form-of-f-x-2-calculate-0-1-ln-1-t-2-t-2-dt-3-calculate-0-1-ln-1-2t-2-t-2-dt-4-calculate-0-1-ln-1-t-2

Question Number 44201 by abdo.msup.com last updated on 23/Sep/18

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

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

2) c a l c u l a t e \int_{0}^{1} \frac{l n (1 + t^{2})}{t^{2}} d t

3) c a l c u l a t e \int_{0}^{1} \frac{l n (1 + 2 t^{2})}{t^{2}} d t

4) c a l c u l a t e \int_{0}^{1} \frac{l n (1 - t^{2})}{t^{2}} d t

Commented by maxmathsup by imad last updated on 25/Sep/18

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

c a s e 1 x > 0 c h a n g e m e n t t \sqrt{x} = u g i v e f^{^{'}} (x) = \int_{0}^{\sqrt{x}} \frac{1}{1 + u^{2}} \frac{d u}{\sqrt{x}}

= \frac{1}{\sqrt{x}} a r c t a n (\sqrt{x}) \Rightarrow f (x) = \int_{0}^{x} \frac{a r c t a n (\sqrt{t})}{\sqrt{t}} d t + c b u t c = f (0) = 0 \Rightarrow

f (x) = \int_{0}^{x} \frac{a c t a n (\sqrt{t})}{\sqrt{t}} d t =_{\sqrt{t} = u} \int_{0}^{\sqrt{x}} \frac{a r c t a n (u)}{u} (2 u) d u = 2 \int_{0}^{\sqrt{x}} a r c t a n u d u

b y p a r t s f (x) = 2 {{[u a r c t a n u]}_{0}^{\sqrt{x}} - \int_{0}^{\sqrt{x}} \frac{u}{1 + u^{2}} d u}

= 2 {\sqrt{x} a r c t a n (\sqrt{x}) - {[\frac{1}{2} l n (1 + u^{2})]}_{0}^{\sqrt{x}}}

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

c a s e 2 x < 0 \Rightarrow f^{^{'}} (x) = \int_{0}^{1} \frac{d t}{1 - (- x) t^{2}} =_{t \sqrt{- x} = u} \int_{0}^{\sqrt{- x}} \frac{1}{1 - u^{2}} \frac{d u}{\sqrt{- x}}

= \frac{1}{2 \sqrt{- x}} {\int_{0}^{\sqrt{- x}} {\frac{1}{1 - u} + \frac{1}{1 + u}} d u} = \frac{1}{2 \sqrt{- x}} {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{\sqrt{- x}} \Rightarrow

f^{^{'}} (x) = \frac{1}{2 \sqrt{- x}} l n ∣ \frac{1 + \sqrt{- x}}{1 - \sqrt{- x}} ∣ \Rightarrow f (x) = \int_{x}^{0} \frac{1}{2 \sqrt{- t}} l n ∣ \frac{1 + \sqrt{- t}}{1 - \sqrt{- t}} ∣ + c

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

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

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

b u t \int_{0}^{\sqrt{- x}} l n (1 + u) d u =_{1 + u = α} \int_{1}^{1 + \sqrt{- x}} l n (α) d α

{[α l n (α) - α]}_{1}^{1 + \sqrt{- x}} = (1 + \sqrt{- x}) l n (1 + \sqrt{- x}) - (1 + \sqrt{- x}) + 1

= (1 + \sqrt{- x}) l n (1 + \sqrt{- x}) - \sqrt{- x} . a l s o

\int_{0}^{\sqrt{- x}} l n (1 - u) d u =_{1 - u = α} - \int_{1}^{1 - \sqrt{- x}} l n (α) d α

= - {[α l n (α) - α]}_{1}^{1 - \sqrt{- x}} = - {(1 - \sqrt{- x}) l n (1 - \sqrt{- x}) - (1 - \sqrt{- x}) + 1}

= - (1 - \sqrt{- x}) l n (1 - \sqrt{- x}) - \sqrt{- x} \Rightarrow

f (x) = (1 + \sqrt{- x}) l n (1 + \sqrt{- x}) + (1 - \sqrt{- x}) l n (1 - \sqrt{- x}) .

Commented by maxmathsup by imad last updated on 25/Sep/18

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

= f (1) = 2 a r c t a n (1) - l n (2) = \frac{2 π}{4} - l n (2) = \frac{π}{2} - l n (2) .

Commented by maxmathsup by imad last updated on 25/Sep/18

\int_{0}^{1} \frac{l n (1 + 2 t^{2})}{t^{2}} d t = f (2) = 2 \sqrt{2} a r c t a n (\sqrt{2}) - l n (3) .

Commented by maxmathsup by imad last updated on 25/Sep/18

4) \int_{0}^{1} \frac{l n (1 - t^{2})}{t^{2}} d t = f (- 1) = 2 l n (2) .

Answered by tanmay.chaudhury50@gmail.com last updated on 25/Sep/18

\frac{d f}{d x} = \int_{0}^{1} \frac{\partial}{\partial x} \times {\frac{l n (1 + {x t}^{2})}{t^{2}}} d t

= \int_{0}^{1} \frac{1}{t^{2}} \times \frac{0 + t^{2}}{1 + {x t}^{2}} d t

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

= \frac{1}{x} \times \frac{1}{\frac{1}{\sqrt{x}}} ∣ {t a n}^{- 1} (\frac{t}{\frac{1}{\sqrt{x_{}}}}) ∣_{0}^{1}

= \frac{1}{\sqrt{x}} {t a n}^{- 1} (\sqrt{x})

d f (x) = \frac{{t a n}^{- 1} (\sqrt{x})}{\sqrt{x}} d x

f (x) = \int \frac{{t a n}^{- 1} (\sqrt{x})}{\sqrt{x}} d x

y^{2} = x

d x = 2 y d y

\int \frac{{t a n}^{- 1} (y) 2 y d y}{y}

= 2 \int {t a n}^{- 1} (y) d y

= 2 {y t a n}^{- 1} (y) - \int \frac{2}{1 + y^{2}} \times y d y

= 2 {y t a n}^{- 1} (y) - l n (1 + y^{2}) + c

= 2 \sqrt{x} {t a n}^{- 1} (\sqrt{x}) - l n (1 + x) + c