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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

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} 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