calculate-0-1-x-1-x-2-1-2-dx-

Question Number 36431 by prof Abdo imad last updated on 02/Jun/18

c a l c u l a t e \int_{0}^{1} \frac{x + 1}{{(x^{2} + 1)}^{2}} d x

Commented by prof Abdo imad last updated on 03/Jun/18

I = \frac{1}{2} \int_{0}^{1} \frac{2 x + 2}{{(x^{2} + 1)}^{2}} d x = \frac{1}{2} \int \frac{2 x}{{(x^{2} + 1)}^{2}} + \int_{0}^{1} \frac{d x}{{(x^{2} + 1)}^{2}}

= {[\frac{- 1}{2 (x^{2} + 1)}]}_{0}^{1} + \int_{0}^{1} \frac{d x}{{(x^{2} + 1)}^{2}} c h a n g e m e n t

x = t a n θ g i v e

\int_{0}^{1} \frac{d x}{{(1 + x^{2})}^{2}} = \int_{0}^{\frac{π}{4}} \frac{(1 + {t a n}^{2} θ)}{{(1 + {t a n}^{2} θ)}^{2}} d θ

= \int_{0}^{\frac{π}{4}} {c o s}^{2} θ d θ = \int_{0}^{\frac{π}{4}} \frac{1 + c o s (2 θ)}{2} d θ

= \frac{π}{8} + \frac{1}{4} {[s i n (2 θ)]}_{0}^{\frac{π}{4}} = \frac{π}{8} + \frac{1}{4} \Rightarrow

I = \frac{1}{2} - \frac{1}{4} + \frac{π}{8} + \frac{1}{4} \Rightarrow

I = \frac{π}{8} + \frac{1}{2} .