find-I-arctan-1-x-dx-and-J-actan-1-x-dx-

Question Number 55995 by maxmathsup by imad last updated on 07/Mar/19

f i n d I = \int a r c t a n (1 - x) d x a n d J = \int a c t a n (1 + x) d x

Commented by maxmathsup by imad last updated on 08/Mar/19

b y p s r t s I = x a r c t a n (1 - x) - \int x \frac{- 1}{1 + {(1 - x)}^{2}} d x

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

\int \frac{x}{x^{2} - 2 x + 2} d x = \frac{1}{2} \int \frac{2 x - 2 + 2}{x^{2} - 2 x + 2} d x = \frac{1}{2} l n ∣ x^{2} - 2 x + 2 ∣ + \int \frac{d x}{x^{2} - 2 x + 2} + c

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

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

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

J = \int a r c t a n (1 + x) d x =_{x = - t} - \int a r c t a n (1 - t) d t

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

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

Answered by MJS last updated on 08/Mar/19

they are standard ?!

\arctan (1 - x) = - \arctan (x - 1)

\int \arctan (x \pm a) d x =

= (x \pm a) \arctan (x \pm a) - \frac{1}{2} \ln ({(x \pm a)}^{2} + 1) + C

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