let-F-x-x-1-x-2-1-arctan-1-t-dt-1-calculate-F-x-x-2-find-lim-x-0-F-x-

Question Number 35682 by prof Abdo imad last updated on 22/May/18

l e t F (x) = \int_{x + 1}^{x^{2} + 1} a r c t a n (1 + t) d t

1) c a l c u l a t e \frac{\partial F}{\partial x} (x)

2) f i n d {l i m}_{x \to 0} F (x) .

Commented by tanmay.chaudhury50@gmail.com last updated on 22/May/18

Commented by abdo mathsup 649 cc last updated on 23/May/18

1) w e h a v e \frac{\partial F}{\partial x} (x) = {(x^{2} + 1)}^{^{'}} a r c t a n (2 + x^{2})

- {(x + 1)}^{^{'}} a r c t a n (2 + x)

= 2 x a r c t a n (3 + x^{2}) - a r c t a n (2 + x)

2) \exists ξ \in] x + 1, x^{2} + 1 [/

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

= (x^{2} - x) a r c t a n (1 + ξ) \Rightarrow {l i m}_{x \to 0} F (x)

= 0 \times a r c t a n (2) = 0 .

Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

\int {t a n}^{- 1} (1 + t) d t

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

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

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

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

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

n o w p u t t i n g t h e u p p e r l i m i t a n d l o w e r l i m i t

u p = (x^{2} + 1) {t a n}^{- 1} (x^{2} + 2) - \frac{1}{2} \times l n (x^{4} + 2^{} x^{2} + 1 +

2 x^{2} + 2 + 2) + {t a n}^{- 1} (x^{2} + 2)

l w = (x + 1) \times {t a n}^{- 1} (x + 2) - \frac{1}{2} \times l n (x^{2} + 2 x + 1 +

2 x + 2 + 2) + {t a n}^{- 1} (x + 2)

r e q u i r e d r e s u l t = \frac{\partial}{\partial x} (u p - l w)

p a g e a r e a n o t s u f f i c i e n t \dots