find-lim-x-0-x-1-2x-1-tarctan-t-2-1-1-1-t-2-2-dt-

Question Number 48177 by Abdo msup. last updated on 20/Nov/18

f i n d {l i m}_{x \to 0} \int_{x + 1}^{2 x + 1} \frac{t a r c t a n (t^{2} + 1)}{1 + {(1 + t^{2})}^{2}} d t

Commented by kaivan.ahmadi last updated on 21/Nov/18

it is zero

Commented by Abdo msup. last updated on 21/Nov/18

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

u^{^{'}} = \frac{t}{1 + {(1 + t^{2})}^{2}} a n d v = a r c t a n (t^{2} + 1) \Rightarrow

A (x) = {[\frac{1}{2} a r c t a n (1 + t^{2}) . a r c t a n (1 + t^{2})]}_{x + 1}^{2 x + 1}

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

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

- A (x) \Rightarrow

A (x) = \frac{1}{4} {{a r c t a n}^{2} (1 + {(2 x + 1)}^{2}) - a r c t a n (1 + {(x + 1)}^{2})} \Rightarrow

{l i m}_{x \to 0} A (x) = \frac{1}{4} {{a r c t a n}^{2} (2) - {a r c t a n}^{2} (2)} = 0 .