∫(x+(√(1+x^2 )) )^n dx


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Question Number 36541 by tanmay.chaudhury50@gmail.com last updated on 03/Jun/18

$\int {(x + \sqrt{1 + x^{2}})}^{n} d x$
Commented by prof Abdo imad last updated on 03/Jun/18
$let put A_n = ∫ (x +(√(1+x^2 )) )^n dx changement x =sh(t) give A_n = ∫ (sht +cht)^n ch dt =∫ { ((e^t −e^(−t) + e^t +e^(−t) )/2)}^n cht dt =∫ e^(nt) ( ((e^t +e^(−t) )/2))dt =(1/2)∫ e^((n+1)t) dt +(1/2)∫ e^((n−1)t) dt = (1/(2(n+1)))e^((n+1)t) +(1/(2(n+1)))e^((n−1)t) +c but e^((n+1)t) =e^((n+1)ln(x+(√(1+x^2 )))) =(x +(√(1+x^2 )))^(n+1) and by?tha same e^((n−1)t) = (x+(√(1+x^2 )))^(n−1) ⇒ A_n = (1/(2(n+1))){ x+(√(1+x^2 )))^(n+1) + (1/(2(n−1))){ x +(√(1+x^2 )))^(n−1) +c if n≠1$
$l e t p u t A_{n} = \int {(x + \sqrt{1 + x^{2}})}^{n} d x c h a n g e m e n t$ $x = s h (t) g i v e$ $A_{n} = \int {(s h t + c h t)}^{n} c h d t$ $= \int {\frac{e^{t} - e^{- t} + e^{t} + e^{- t}}{2}}^{n} c h t d t$ $= \int e^{n t} (\frac{e^{t} + e^{- t}}{2}) d t$ $= \frac{1}{2} \int e^{(n + 1) t} d t + \frac{1}{2} \int e^{(n - 1) t} d t$ $= \frac{1}{2 (n + 1)} e^{(n + 1) t} + \frac{1}{2 (n + 1)} e^{(n - 1) t} + c$ $b u t e^{(n + 1) t} = e^{(n + 1) l n (x + \sqrt{1 + x^{2}})}$ $= {(x + \sqrt{1 + x^{2}})}^{n + 1} a n d b y ? t h a s a m e$ $e^{(n - 1) t} = {(x + \sqrt{1 + x^{2}})}^{n - 1} \Rightarrow$ $A_{n} = \frac{1}{2 (n + 1)} {{x + \sqrt{1 + x^{2}})}^{n + 1}$ $+ \frac{1}{2 (n - 1)} {{x + \sqrt{1 + x^{2}})}^{n - 1} + c i f n \neq 1$
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