∫_0 ^a ((a^2 −x^2 )/((a^2 +x^2 )^2 )) dx ?


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Question Number 94312 by i jagooll last updated on 18/May/20

$\underset{0}{\int^{a}} \frac{a^{2} - x^{2}}{{(a^{2} + x^{2})}^{2}} d x ?$
Commented by mathmax by abdo last updated on 18/May/20
$let I =∫_0 ^a ((a^2 −x^2 )/((a^2 +x^2 )^2 ))dx ⇒I =_(x=at) ∫_0 ^1 ((a^2 −a^2 t^2 )/(a^4 (1+t^2 )^2 )) ×adt =(1/a) ∫_0 ^1 ((1−t^2 )/((1+t^2 )^2 ))dt ⇒aI =∫_0 ^1 ((1+t^2 −2t^2 )/((1+t^2 )^2 ))dt =∫_0 ^1 (dt/(1+t^2 )) −2 ∫_0 ^1 (t^2 /((1+t^2 )^2 ))dt we have ∫_0 ^1 (dt/(1+t^2 )) =[arctant]_0 ^1 =(π/4) ∫_0 ^1 (t^2 /((1+t^2 )^2 ))dt =∫_0 ^1 ((1+t^2 −1)/((1+t^2 )^2 ))dt =∫_0 ^1 (dt/(1+t^2 ))−∫_0 ^1 (dt/((1+t^2 )^2 )) =(π/4)−∫_0 ^1 (dt/((1+t^2 )^2 )) ∫_0 ^1 (dt/((1+t^2 )^2 )) =_(t=tanu) ∫_0 ^(π/4) ((1+tan^2 u)/((1+tan^2 u)^2 ))du =∫_0 ^(π/4) (du/(1+tan^2 u)) =∫_0 ^(π/4) cos^2 udu =(1/2)∫_0 ^(π/2) (1+cos(2u))du =(π/4) +(1/4)[sin(2u)]_0 ^(π/4) =(π/4)+(1/4) ⇒aI =(π/4) −2{(π/4)−(π/4)−(1/4)} =(π/4) +(1/2) ⇒I =(1/a)((π/4)+(1/2)) with a≠0 for a=0 I is not defined !$
$l e t I = \int_{0}^{a} \frac{a^{2} - x^{2}}{{(a^{2} + x^{2})}^{2}} d x \Rightarrow I =_{x = a t} \int_{0}^{1} \frac{a^{2} - a^{2} t^{2}}{a^{4} {(1 + t^{2})}^{2}} \times a d t$ $= \frac{1}{a} \int_{0}^{1} \frac{1 - t^{2}}{{(1 + t^{2})}^{2}} d t \Rightarrow a I = \int_{0}^{1} \frac{1 + t^{2} - 2 t^{2}}{{(1 + t^{2})}^{2}} d t$ $= \int_{0}^{1} \frac{d t}{1 + t^{2}} - 2 \int_{0}^{1} \frac{t^{2}}{{(1 + t^{2})}^{2}} d t w e h a v e \int_{0}^{1} \frac{d t}{1 + t^{2}} = {[a r c t a n t]}_{0}^{1} = \frac{π}{4}$ $\int_{0}^{1} \frac{t^{2}}{{(1 + t^{2})}^{2}} d t = \int_{0}^{1} \frac{1 + t^{2} - 1}{{(1 + t^{2})}^{2}} d t = \int_{0}^{1} \frac{d t}{1 + t^{2}} - \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}} = \frac{π}{4} - \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}}$ $\int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}} =_{t = t a n u} \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{2} u}{{(1 + {t a n}^{2} u)}^{2}} d u = \int_{0}^{\frac{π}{4}} \frac{d u}{1 + {t a n}^{2} u}$ $= \int_{0}^{\frac{π}{4}} {c o s}^{2} u d u = \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (2 u)) d u = \frac{π}{4} + \frac{1}{4} {[s i n (2 u)]}_{0}^{\frac{π}{4}}$ $= \frac{π}{4} + \frac{1}{4} \Rightarrow a I = \frac{π}{4} - 2 {\frac{π}{4} - \frac{π}{4} - \frac{1}{4}}$ $= \frac{π}{4} + \frac{1}{2} \Rightarrow I = \frac{1}{a} (\frac{π}{4} + \frac{1}{2}) w i t h a \neq 0$ $f o r a = 0 I i s n o t d e f i n e d!$
Commented by mathmax by abdo last updated on 18/May/20
$error of typo ∫_0 ^1 (dt/((1+t^2 )^2 )) =(π/8) +(1/4) ⇒aI =(π/4)−2{(π/4)−(π/8)−(1/4)} =(π/4)−(π/2) +(π/4) +(1/2) =(1/2) ⇒ I =(1/(2a))$
$e r r o r o f t y p o \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{2}} = \frac{π}{8} + \frac{1}{4} \Rightarrow a I = \frac{π}{4} - 2 {\frac{π}{4} - \frac{π}{8} - \frac{1}{4}}$ $= \frac{π}{4} - \frac{π}{2} + \frac{π}{4} + \frac{1}{2} = \frac{1}{2} \Rightarrow I = \frac{1}{2 a}$
Commented by i jagooll last updated on 18/May/20

$yes . ===$
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