1-1-xdx-x-2-1-2-

Question Number 26486 by A1B1C1D1 last updated on 26/Dec/17

\int_{- 1}^{1} \frac{xdx}{{(x^{2} + 1)}^{2}}

Commented by kaivan.ahmadi last updated on 26/Dec/17

u = x^{2} + 1 \Rightarrow du = 2 xdx

\Rightarrow \frac{1}{2} \int \frac{du}{u^{2}} = 1 / 2 (- \frac{1}{u}) = - 1 / 2 (\frac{1}{x^{2} + 1}) ∣_{- 1}^{1} =

- 1 / 2 (\frac{1}{2} - \frac{1}{2}) = 0

Commented by A1B1C1D1 last updated on 26/Dec/17

Thanks .

Commented by mrW1 last updated on 26/Dec/17

f o r a n y f (x) :

i f f (- x) = - f (x)

\int_{- a}^{a} f (x) d x = \int_{- a}^{0} f (x) d x + \int_{0}^{a} f (x) d x

= - \int_{0}^{a} f (x) d x + \int_{0}^{a} f (x) d x

= 0

Commented by A1B1C1D1 last updated on 26/Dec/17

Thanks

Commented by abdo imad last updated on 26/Dec/17

i n g e n e r a l i f f i s i n t e g r a b l e a n d o d d (f (- x) = - f (x))

o n [- a, a] \Rightarrow \int_{- a}^{a} f (x) d x = 0

Answered by jota@ last updated on 26/Dec/17

= 0 . T h e f u n c t i o n i s i m p a r .

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