Question Number 57419 by Abdo msup. last updated on 03/Apr/19
$${find}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}+\mathrm{1}\right)\:{ln}\left({x}+\sqrt{\left.\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx}\right. \\ $$
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).](https://www.tinkutara.com/question/Q57569.png)
$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left({x}+\mathrm{1}\right){ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right){dx}\:\:{by}\:{parts}\:{u}^{'} \:={x}+\mathrm{1}\:{and}\:{v}\:={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${A}\:=\left[\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{x}\right){ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{x}\right)\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{1}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} \:+\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}^{\mathrm{2}} \:+\mathrm{1}+\mathrm{2}{x}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}{dx}\:\:+\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:=_{{x}\:={sh}\left({t}\right)} \:\:\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \:{ch}\left({t}\right){ch}\left({t}\right){dt} \\ $$$$=\int_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\:{dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left[{sh}\left(\mathrm{2}{t}\right)\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{4}}\left[\:\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\right]_{\mathrm{0}} ^{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:+\frac{\mathrm{1}}{\mathrm{8}}\left\{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:−\frac{\mathrm{1}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)^{\mathrm{2}} }\right\}\:. \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx}\:=\left[\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\sqrt{\mathrm{2}}−\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\left[{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} \:={ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:\Rightarrow\right. \\ $$$${A}\:=\frac{\mathrm{3}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{16}}\left\{\:\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right\}−\sqrt{\mathrm{2}}\:+\mathrm{1}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right) \\ $$$${A}\:=\frac{\mathrm{7}}{\mathrm{4}}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:−\frac{\mathrm{1}}{\mathrm{16}}\left\{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\right\}\:+\mathrm{1}−\sqrt{\mathrm{2}}. \\ $$