find-ln-x-x-2-1-x-2-1-dx-2-calculate-2-5-ln-x-x-2-1-x-2-1-dx-

Question Number 39833 by math khazana by abdo last updated on 12/Jul/18

f i n d \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x

2) c a l c u l a t e \int_{2}^{5} \frac{l n (x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}} d x

Commented by abdo mathsup 649 cc last updated on 12/Jul/18

l e t I = \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x l e t i n t e g r a t e b y p a r t s

u^{^{'}} = \frac{1}{\sqrt{x^{2} - 1}} a n d v (x) = l n (x + \sqrt{x^{2} - 1})

I = {l n}^{2} (x + \sqrt{x^{2} - 1}) - \int \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x

= {l n}^{2} (x + \sqrt{x^{2} - 1}) - I \Rightarrow

2 I = {l n}^{2} (x + \sqrt{x^{2} - 1}) \Rightarrow I = \frac{1}{2} {l n}^{2} (x + \sqrt{x^{2} - 1})

2) \int_{2}^{5} \frac{l n (x + \sqrt{x^{2} - 1})}{\sqrt{x^{2} - 1}} d x = {[\frac{1}{2} {l n}^{2} (x + \sqrt{x^{2} - 1})]}_{2}^{5}

= \frac{1}{2} {{l n}^{2} (5 + \sqrt{24}) - {l n}^{2} (2 + \sqrt{3})}

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

t = x + \sqrt{x^{2} - 1}

\frac{d t}{d x} = 1 + \frac{2 x}{2 \sqrt{x^{2} - 1}} = \frac{x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}}

\int l n t d t

t l n t - t + c

(x + \sqrt{x^{2} - 1}) l n (x + \sqrt{x^{2} - 1}) - (x + \sqrt{x^{2} - 1}) + c

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Jul/18

t = x + \sqrt{x^{2} - 1}

\frac{d t}{d x} = 1 + \frac{2 x}{2 \sqrt{x^{2} - 1}} = \frac{x + \sqrt{x^{2} - 1}}{\sqrt{x^{2} - 1}}

\int l n t d t

t l n t - t + c

(x + \sqrt{x^{2} - 1}) l n (x + \sqrt{x^{2} - 1}) - (x + \sqrt{x^{2} - 1}) + c