1-find-dx-x-1-x-2-2-calculate-0-1-dx-x-1-x-2-

Question Number 64418 by turbo msup by abdo last updated on 17/Jul/19

$$\left.\mathrm{1}\right){find}\:\int\:\:\:\frac{{dx}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$

Commented by Prithwish sen last updated on 17/Jul/19

a)∫((√(x^2 +1))−x)dx = (x/2)(√(x^2 +1))+(1/2)ln∣x+(√(x^2 +1)) ∣ −(x^2 /2) +C b) ((x/2)(√(x^2 +1)) + (1/2)ln∣x+(√(x^2 +1))∣−(x^2 /2) )_0 ^1 = ((1/( (√2))) +(1/2)ln∣1+(√2)∣−(1/2))

$$\left.\mathrm{a}\right)\int\left(\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}−\mathrm{x}\right)\mathrm{dx}\:=\: \\ $$$$\frac{\mathrm{x}}{\mathrm{2}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:\mid\:−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:+\mathrm{C} \\ $$$$\left.\mathrm{b}\right)\:\left(\frac{\mathrm{x}}{\mathrm{2}}\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}\mid−\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:\right)_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\:\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{1}+\sqrt{\mathrm{2}}\mid−\frac{\mathrm{1}}{\mathrm{2}}\right)\: \\ $$

Commented by mathmax by abdo last updated on 18/Jul/19

1) let I =∫ (dx/(x+(√(1+x^2 )))) ⇒ I =∫ ((x−(√(1+x^2 )))/(x^2 −1−x^2 ))dx = ∫ (√(1+x^2 ))dx−∫ xdx =−(x^2 /2) +∫ (√(1+x^2 ))dx +c changement x =sh(t) give ∫(√(1+x^2 ))dx =∫ ch(t)cht dt =∫ ch^2 t dt =∫ ((1+ch(2t))/2)dt =(t/2) +(1/4)sh(2t) =(t/2) +(1/2)sh(t)ch(t)=(t/2) +(1/2)x(√(1+x^2 )) =(1/2)argsh(x) +((x(√(1+x^2 )))/2) =(1/2)ln(x+(√(1+x^2 ))) +((x(√(1+x^2 )))/2) ⇒ I =(1/2)ln(x+(√(1+x^2 )))+((x(√(1+x^2 )))/2) −(x^2 /2) +c .

$$\left.\mathrm{1}\right)\:{let}\:\:\:{I}\:=\int\:\:\frac{{dx}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:\Rightarrow\:{I}\:=\int\:\frac{{x}−\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} }{dx} \\ $$$$=\:\int\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}−\int\:{xdx}\:=−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+\int\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:+{c} \\ $$$${changement}\:{x}\:={sh}\left({t}\right)\:{give} \\ $$$$\int\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=\int\:{ch}\left({t}\right){cht}\:{dt}\:=\int\:{ch}^{\mathrm{2}} {t}\:{dt}\:=\int\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{argsh}\left({x}\right)\:+\frac{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}}\:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\frac{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}}\:−\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\:+{c}\:. \\ $$

Commented by mathmax by abdo last updated on 18/Jul/19

2) ∫_0 ^1 (dx/(x+(√(1+x^2 )))) =(1/2)[ln(x+(√(1+x^2 ))) +x(√(1+x^2 )) −x^2 ]_0 ^1 =(1/2){ln(1+(√2))+(√2)−1}

$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dx}}{{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−{x}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left\{{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)+\sqrt{\mathrm{2}}−\mathrm{1}\right\} \\ $$

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