find ∫ (1+(1/t^2 ))arctan(t−(1/t))dt .


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Question Number 42488 by maxmathsup by imad last updated on 26/Aug/18

$f i n d \int (1 + \frac{1}{t^{2}}) a r c t a n (t - \frac{1}{t}) d t .$
Commented by maxmathsup by imad last updated on 27/Aug/18

$b y p a r t s u^{^{'}} = 1 + \frac{1}{t^{2}} a n d v = a r c t a n (t - \frac{1}{t}) \Rightarrow$ $I = (t - \frac{1}{t}) a r c t a n (t - \frac{1}{t}) - \int (t - \frac{1}{t}) \frac{1 + \frac{1}{t^{2}}}{1 + {(t - \frac{1}{t})}^{2}} d t$ $= (t - \frac{1}{t}) a r c t a n (t - \frac{1}{t}) - \int \frac{(t - 1) (t^{2} + 1)}{t^{3} (1 + t^{2} - 2 + \frac{1}{t^{2}})} d t$ $= (t - \frac{1}{t}) a r c t a n (t - \frac{1}{t}) - \int \frac{(t - 1) (t^{2} + 1)}{t (t^{2} + t^{4} - 2 t^{2} + 1)} d t b u t$ $\int \frac{(t - 1) (t^{2} + 1)}{t (t^{2} + t^{4} - 2 t^{2} + 1)} d t = \int \frac{t^{3} + t - t^{2} - 1}{t^{3} + t^{5} - 2 t^{3} + 1} d t$ $= \int \frac{t^{3} - t^{2} + t - 1}{t^{5} - t^{3} + 1} d t l e t d e c o m p o s e F (t) = \frac{t^{3} - t^{2} + t - 1}{t^{5} - t^{3} + 1}$ $t h e r o o t s o f t^{5} - t^{3} + 1 a r e$ $t_{1} \sim 0, 959 + 0, 4284 i (c o m p l e x)$ $t_{2} \sim 0, 959 - 0, 4284 i (c o m p l e x)$ $t_{3} \sim - 1, 2365 (r e e l)$ $t_{4} \sim - 0, 3408 + 0, 7854 i (c o m p l e x)$ $t_{5} \sim - 0, 3408 - 0, 7854 i (c o m l e x) \Rightarrow$ $F (t) \sim \frac{t^{3} - t^{2} + t - 1}{(t - t_{1}) (t - {\bar{t}}_{1}) (t - t_{3}) (t - t_{4}) (t - {\bar{t}}_{4})}$ $\sim \frac{t^{3} - t^{2} + t - 1}{(t - t_{3}) (t^{2} - 2 R e (t_{1}) t + 1) (t^{2} - 2 R e (t_{4}) t + 1)}$ $\sim \frac{a}{t - t_{3}} + \frac{b t + c}{t^{2} - 2 R e (t_{1}) t + 1} + \frac{d t + e}{t^{2} - 2 R e (t_{4}) t + 1} \Rightarrow$ $\int F (t) d t \sim \int \frac{a d t}{t - t_{3}} + \int \frac{b t + c}{t^{2} - 2 R e (t_{1}) t + 1} + \int \frac{d t + e}{t^{2} - 2 R e (t_{4}) t + 1} + c . . .$ $. . . b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/Aug/18

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