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Question Number 27853 by Poojadarshini94 last updated on 15/Jan/18

Commented by abdo imad last updated on 17/Jan/18

let put f(x)= ln(x+(√(x^2 +a^2 )))  we have   f′(x)=^(  )   ((1 +((2x)/(2(√(x^2 +a^2 )))))/(x+(√(x^2  +a^2 )))) =  ((x +(√(x^2 +a^2 )))/(√(x^2 +a^2 ))) .  (1/(x+(√(x^2 +a^2 ))))  = (1/(√(x^2 +a^2 ))) ⇒    ∫   (dx/(√(x^2 +a^2 ))) =  ln(x+(√(x^2 +a^2 ))) +k  another method you can use the ch. x=asht ...

$${let}\:{put}\:{f}\left({x}\right)=\:{ln}\left({x}+\sqrt{\left.{x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)}\:\:{we}\:{have}\:\right. \\ $$$${f}'\left({x}\right)=^{\:\:} \:\:\frac{\mathrm{1}\:+\frac{\mathrm{2}{x}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}}{{x}+\sqrt{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }}\:=\:\:\frac{{x}\:+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\:.\:\:\frac{\mathrm{1}}{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\:\Rightarrow\:\:\:\:\int\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}\:=\:\:{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)\:+{k} \\ $$$${another}\:{method}\:{you}\:{can}\:{use}\:{the}\:{ch}.\:{x}={asht}\:... \\ $$

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