1-1-x-2-2-

Question Number 93473 by mashallah last updated on 13/May/20

\int 1 / {(1 + x^{2})}^{2}

Commented by abdomathmax last updated on 15/May/20

A = \int \frac{d x}{{(1 + x^{2})}^{2}} w e d o t h e c h a n g e m e n t x = t a n t \Rightarrow

A = \int \frac{(1 + {t a n}^{2} t) d t}{{(1 + {t a n}^{2} t)}^{2}} = \int {c o s}^{2} t d t = \frac{1}{2} \int (1 + c o s (2 t)) d t

= \frac{t}{2} + \frac{1}{4} s i n (2 t) + C

= \frac{a r c t a n x}{2} + \frac{1}{4} \times \frac{t a n t}{1 + {t a n}^{2} t} + C

= \frac{a r c t a n x}{2} + \frac{1}{4} \frac{x}{1 + x^{2}} + C

Answered by Rio Michael last updated on 13/May/20

\int \frac{1}{{(1 + x^{2})}^{2}} d x

changement x = \tan θ \Rightarrow \frac{d x}{d θ} = \sec^{2} θ

\Rightarrow \int \frac{1}{{(1 + x^{2})}^{2}} d x = \int \frac{1}{{(1 + \tan^{2} θ)}^{2}} \sec^{2} θ d θ = \int \frac{1}{\sec^{2} θ} d θ

\Rightarrow \int \cos^{2} θ d θ = \frac{1}{2} \int (1 + \cos 2 θ) d θ = \frac{1}{2} (θ + \frac{\sin 2 θ}{2}) + C