f is a function positive and C^1 1) find ∫ (f^′ /(2(√f)(√(1+f))))dx 2)let A_n = ∫_0 ^1 (x^(n/2) /(x(√(1+x^n )))) calculate A_n and lim_(n→+∞) A


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Question Number 38123 by maxmathsup by imad last updated on 22/Jun/18

$f i s a f u n c t i o n p o s i t i v e a n d C^{1}$ $1) f i n d \int \frac{f^{^{'}}}{2 \sqrt{f} \sqrt{1 + f}} d x$ $2) l e t A_{n} = \int_{0}^{1} \frac{x^{\frac{n}{2}}}{x \sqrt{1 + x^{n}}}$ $c a l c u l a t e A_{n} a n d {l i m}_{n \to + \infty} A_{n}$
Commented by prof Abdo imad last updated on 22/Jun/18
$changement (√f)=u (u function!) give ∫ (f^′ /(2(√f)(√(1+f))))dx=∫ ((2u^′ u)/(2u(√(1+u^2 ))))dx =∫ (u^′ /(√(1+u^2 )))dx=ln(u +(√(1+u^2 ))) +c =ln((√f) +(√(1+f))) +c 2) we have A_n = ∫_0 ^1 (x^(n/2) /(x(√(1+x^n ))))dx changement x^n =t give x=t^(1/n) and A_n = ∫_0 ^1 (t^(1/2) /(t^((1/n) ) (√(1+t)))) (1/n)t^((1/n) −1) dt =(1/n) ∫_0 ^1 (dt/((√t)(√(1+t)))) =_((√t)=u) (1/n)∫_0 ^1 ((2udu)/(u(√(1+u^2 )))) = (2/n)[ln(u +(√(1+u^2 )))]_0 ^1 =(2/n){ ln(1+(√2))}$
$c h a n g e m e n t \sqrt{f} = u (u f u n c t i o n!) g i v e$ $\int \frac{f^{^{'}}}{2 \sqrt{f} \sqrt{1 + f}} d x = \int \frac{2 u^{^{'}} u}{2 u \sqrt{1 + u^{2}}} d x$ $= \int \frac{u^{^{'}}}{\sqrt{1 + u^{2}}} d x = l n (u + \sqrt{1 + u^{2}}) + c$ $= l n (\sqrt{f} + \sqrt{1 + f}) + c$ $2) w e h a v e A_{n} = \int_{0}^{1} \frac{x^{\frac{n}{2}}}{x \sqrt{1 + x^{n}}} d x$ $c h a n g e m e n t x^{n} = t g i v e x = t^{\frac{1}{n}} a n d$ $A_{n} = \int_{0}^{1} \frac{t^{\frac{1}{2}}}{t^{\frac{1}{n}} \sqrt{1 + t}} \frac{1}{n} t^{\frac{1}{n} - 1} d t$ $= \frac{1}{n} \int_{0}^{1} \frac{d t}{\sqrt{t} \sqrt{1 + t}} =_{\sqrt{t} = u} \frac{1}{n} \int_{0}^{1} \frac{2 u d u}{u \sqrt{1 + u^{2}}}$ $= \frac{2}{n} {[l n (u + \sqrt{1 + u^{2}})]}_{0}^{1}$ $= \frac{2}{n} {l n (1 + \sqrt{2})}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

	$1) \int \frac{\frac{d f}{d x}}{2 \sqrt{f} \sqrt{1 + f}} d x$ $\int \frac{d f}{2 \sqrt{f} \sqrt{1 + f}}$ $t^{2} = 1 + f$ $\int \frac{2 t d t}{2 \sqrt{t^{2} - 1} \times t}$ $\int \frac{d t}{\sqrt{t^{2} - 1}} u s e f o r m u l a \int \frac{d x}{\sqrt{x^{2} - a^{2}}}$ $l n (t + \sqrt{t^{2} - 1}) + c$ $l n (\sqrt{1 + f} + \sqrt{f}) + c$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18
	$x^n +1=t^2 2tdt =nx^(n−1) dx 2tdt=nx^n (dx/x) ((2tdt)/(n(t^2 −1)))=(dx/x) ∫_1 ^((√2) ) ((2tdt)/(n(t^2 −1)))×(((√(t^2 −1)) )/t) =(2/n)∫_1 ^(√2) (√(t^2 −1)) dt (1/n)∣(t/2)(√(t^2 −1)) −(1/2)ln(t+(√(t^2 −1))) ∣_1 ^((√2) ) (1/n){(((√2) )/2)−(1/2)ln((√2) +1)} =(1/n){(((√2) )/2)−(1/2)ln((√2) +1)} when n→∞ value of intregation is 0 use formula∫(√(x^2 −a^2 )) dx$
	$x^{n} + 1 = t^{2} 2 t d t = {n x}^{n - 1} d x$ $2 t d t = {n x}^{n} \frac{d x}{x}$ $\frac{2 t d t}{n (t^{2} - 1)} = \frac{d x}{x}$ $\int_{1}^{\sqrt{2}} \frac{2 t d t}{n (t^{2} - 1)} \times \frac{\sqrt{t^{2} - 1}}{t}$ $= \frac{2}{n} \int_{1}^{\sqrt{2}} \sqrt{t^{2} - 1} d t$ $\frac{1}{n} ∣ \frac{t}{2} \sqrt{t^{2} - 1} - \frac{1}{2} l n (t + \sqrt{t^{2} - 1)} ∣_{1}^{\sqrt{2}}$ $\frac{1}{n} {\frac{\sqrt{2}}{2} - \frac{1}{2} l n (\sqrt{2} + 1)}$ $= \frac{1}{n} {\frac{\sqrt{2}}{2} - \frac{1}{2} l n (\sqrt{2} + 1)}$ $w h e n n \to \infty v a l u e o f i n t r e g a t i o n i s 0$ $u s e f o r m u l a \int \sqrt{x^{2} - a^{2}} d x$
	Commented by tanmay.chaudhury50@gmail.com last updated on 22/Jun/18

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