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dx-2-4-




Question Number 86034 by ar247 last updated on 26/Mar/20
∫(dx/( (√(×^2 +4))))
$$\int\frac{{dx}}{\:\sqrt{×^{\mathrm{2}} +\mathrm{4}}} \\ $$
Answered by TANMAY PANACEA. last updated on 26/Mar/20
x=2tana  (dx/da)=2sec^2 a  ∫((2sec^2 ada)/(2seca))  ∫seca da=ln(seca+tana)+c  ln((x/2)+(√(1+(x^2 /4))) )+c  ln(x+(√(x^2 +4)) )−ln2+c  ln(x+(√(x^2 +4)) )+C
$${x}=\mathrm{2}{tana}\:\:\frac{{dx}}{{da}}=\mathrm{2}{sec}^{\mathrm{2}} {a} \\ $$$$\int\frac{\mathrm{2}{sec}^{\mathrm{2}} {ada}}{\mathrm{2}{seca}} \\ $$$$\int{seca}\:{da}={ln}\left({seca}+{tana}\right)+{c} \\ $$$${ln}\left(\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}}\:\right)+{c} \\ $$$${ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:\right)−{ln}\mathrm{2}+{c} \\ $$$${ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:\right)+{C} \\ $$
Answered by Rio Michael last updated on 26/Mar/20
 ∫(dx/( (√(x^2  + 4)))) = arsinh ((x/2)) + k
$$\:\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{4}}}\:=\:\mathrm{arsinh}\:\left(\frac{{x}}{\mathrm{2}}\right)\:+\:{k}\: \\ $$

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