𝛗=∫_0 ^( 1) ((ln(x^2 +1))/x^2 )dx f(a)=∫_0 ^( 1) ((log(ax^2 +1))/x^2 )dx f ′(a)=∫_0 ^( 1) (x^2 /((ax^2 +1)x^2 ))dx=(1/a)∫_0 ^( 1) (dx/(x^2 +((1/( (√a))))^2 )) =((√a)/a)[tan^(−1) (x(√a) )]_0 ^1 =((√a)/a)tan^(−1) ((√a)) f(a)=^((√a) =u) ∫2tan^(−1) (u)du =2{u.tan^(−1) (u)−∫(u/(1+u^2 ))du}+C =2(√a) tan^(−1) ((√a) )−ln(1+a)+C f(0)=0=0+C⇒C=0 f(1)=𝛗=2((π/4))−ln(2)=(π/2)−ln(2)


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Question Number 136644 by mnjuly1970 last updated on 24/Mar/21
$𝛗=∫_0 ^( 1) ((ln(x^2 +1))/x^2 )dx f(a)=∫_0 ^( 1) ((log(ax^2 +1))/x^2 )dx f ′(a)=∫_0 ^( 1) (x^2 /((ax^2 +1)x^2 ))dx=(1/a)∫_0 ^( 1) (dx/(x^2 +((1/( (√a))))^2 )) =((√a)/a)[tan^(−1) (x(√a) )]_0 ^1 =((√a)/a)tan^(−1) ((√a)) f(a)=^((√a) =u) ∫2tan^(−1) (u)du =2{u.tan^(−1) (u)−∫(u/(1+u^2 ))du}+C =2(√a) tan^(−1) ((√a) )−ln(1+a)+C f(0)=0=0+C⇒C=0 f(1)=𝛗=2((π/4))−ln(2)=(π/2)−ln(2)$
$ϕ = \int_{0}^{1} \frac{l n (x^{2} + 1)}{x^{2}} d x$ $f (a) = \int_{0}^{1} \frac{l o g ({a x}^{2} + 1)}{x^{2}} d x$ $f^{'} (a) = \int_{0}^{1} \frac{x^{2}}{({a x}^{2} + 1) x^{2}} d x = \frac{1}{a} \int_{0}^{1} \frac{d x}{x^{2} + {(\frac{1}{\sqrt{a}})}^{2}}$ $= \frac{\sqrt{a}}{a} {[{t a n}^{- 1} (x \sqrt{a})]}_{0}^{1} = \frac{\sqrt{a}}{a} {t a n}^{- 1} (\sqrt{a})$ $f (a) \overset{\sqrt{a} = u}{=} \int 2 {t a n}^{- 1} (u) d u$ $= 2 {u . {t a n}^{- 1} (u) - \int \frac{u}{1 + u^{2}} d u} + C$ $= 2 \sqrt{a} {t a n}^{- 1} (\sqrt{a}) - l n (1 + a) + C$ $f (0) = 0 = 0 + C \Rightarrow C = 0$ $f (1) = ϕ = 2 (\frac{π}{4}) - l n (2) = \frac{π}{2} - l n (2)$
	Answered by mnjuly1970 last updated on 24/Mar/21

	$m e t h o d 2 :$ $ϕ \overset{i . b . p}{=} {[\frac{- 1}{x} l n (x^{2} + 1)]}_{0}^{1} + \int_{0}^{1} \frac{2 x}{x (1 + x^{2})} d x$ $= - l o g (2) + \frac{π}{2} . .$
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