find ∫_0 ^1 (x+1) ln(x+(√(1+x^2 )))dx


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Question Number 57419 by Abdo msup. last updated on 03/Apr/19

$f i n d \int_{0}^{1} (x + 1) l n (x + \sqrt{1 + x^{2})} d x$
Commented by maxmathsup by imad last updated on 07/Apr/19
$let A =∫_0 ^1 (x+1)ln(x+(√(1+x^2 )))dx by parts u^′ =x+1 and v =ln(x+(√(1+x^2 ))) ⇒ A =[((x^2 /2) +x)ln(x+(√(1+x^2 )))]_0 ^1 −∫_0 ^1 ((x^2 /2) +x)(dx/(√(1+x^2 ))) =(3/2)ln(1+(√2)) −(1/2) ∫_0 ^1 ((x^2 +2x)/(√(1+x^2 ))) dx ∫_0 ^1 ((x^2 +2x)/(√(1+x^2 )))dx =∫_0 ^1 ((x^2 +1+2x−1)/(√(1+x^2 )))dx =∫_0 ^1 (√(x^2 +1))dx +2 ∫_0 ^1 (x/(√(1+x^2 )))dx −∫_0 ^1 (dx/(√(1+x^2 ))) ∫_0 ^1 (√(1+x^2 ))dx =_(x =sh(t)) ∫_0 ^(ln(1+(√2))) ch(t)ch(t)dt =∫_0 ^(ln(1+(√2))) ((1+ch(2t))/2) dt =(1/2)ln(1+(√2)) +(1/4)[sh(2t)]_0 ^(ln(1+(√2))) =(1/2)ln(1+(√2)) +(1/4)[ ((e^(2t) −e^(−2t) )/2)]_0 ^(ln(1+(√2))) =(1/2)ln(1+(√2)) +(1/8){(1+(√2))^2 −(1/((1+(√2))^2 ))} . ∫_0 ^1 (x/(√(1+x^2 ))) dx =[(√(1+x^2 ))]_0 ^1 =(√2)−1 ∫_0 ^1 (dx/(√(1+x^2 ))) =[ln(x+(√(1+x^2 ))]_0 ^1 =ln(1+(√2)) ⇒ A =(3/2)ln(1+(√2)) −(1/4)ln(1+(√2))−(1/(16)){ 3+2(√2)−(1/(3+2(√2)))}−(√2) +1 +(1/2)ln(1+(√2)) A =(7/4)ln(1+(√2)) −(1/(16)){3+2(√2)−(1/(3+2(√2)))} +1−(√2).$
$l e t A = \int_{0}^{1} (x + 1) l n (x + \sqrt{1 + x^{2}}) d x b y p a r t s u^{^{'}} = x + 1 a n d v = l n (x + \sqrt{1 + x^{2}}) \Rightarrow$ $A = {[(\frac{x^{2}}{2} + x) l n (x + \sqrt{1 + x^{2}})]}_{0}^{1} - \int_{0}^{1} (\frac{x^{2}}{2} + x) \frac{d x}{\sqrt{1 + x^{2}}}$ $= \frac{3}{2} l n (1 + \sqrt{2}) - \frac{1}{2} \int_{0}^{1} \frac{x^{2} + 2 x}{\sqrt{1 + x^{2}}} d x$ $\int_{0}^{1} \frac{x^{2} + 2 x}{\sqrt{1 + x^{2}}} d x = \int_{0}^{1} \frac{x^{2} + 1 + 2 x - 1}{\sqrt{1 + x^{2}}} d x = \int_{0}^{1} \sqrt{x^{2} + 1} d x + 2 \int_{0}^{1} \frac{x}{\sqrt{1 + x^{2}}} d x - \int_{0}^{1} \frac{d x}{\sqrt{1 + x^{2}}}$ $\int_{0}^{1} \sqrt{1 + x^{2}} d x =_{x = s h (t)} \int_{0}^{l n (1 + \sqrt{2})} c h (t) c h (t) d t$ $= \int_{0}^{l n (1 + \sqrt{2})} \frac{1 + c h (2 t)}{2} d t = \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{4} {[s h (2 t)]}_{0}^{l n (1 + \sqrt{2})}$ $= \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{4} {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{0}^{l n (1 + \sqrt{2})}$ $= \frac{1}{2} l n (1 + \sqrt{2}) + \frac{1}{8} {{(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}} .$ $\int_{0}^{1} \frac{x}{\sqrt{1 + x^{2}}} d x = {[\sqrt{1 + x^{2}}]}_{0}^{1} = \sqrt{2} - 1$ $\int_{0}^{1} \frac{d x}{\sqrt{1 + x^{2}}} = [l n {(x + \sqrt{1 + x^{2}}]}_{0}^{1} = l n (1 + \sqrt{2}) \Rightarrow$ $A = \frac{3}{2} l n (1 + \sqrt{2}) - \frac{1}{4} l n (1 + \sqrt{2}) - \frac{1}{16} {3 + 2 \sqrt{2} - \frac{1}{3 + 2 \sqrt{2}}} - \sqrt{2} + 1 + \frac{1}{2} l n (1 + \sqrt{2})$ $A = \frac{7}{4} l n (1 + \sqrt{2}) - \frac{1}{16} {3 + 2 \sqrt{2} - \frac{1}{3 + 2 \sqrt{2}}} + 1 - \sqrt{2} .$
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