Question-190318

Question Number 190318 by Rupesh123 last updated on 31/Mar/23

Answered by som(math1967) last updated on 31/Mar/23

let 2x−x^2 −1=t^2 (2−2x)dx=2tdt ∫((2tdt)/t)=2t+C=2(√(2x−x^2 −1))+C

$$\:{let}\:\mathrm{2}{x}−{x}^{\mathrm{2}} −\mathrm{1}={t}^{\mathrm{2}} \\ $$$$\:\left(\mathrm{2}−\mathrm{2}{x}\right){dx}=\mathrm{2}{tdt} \\ $$$$\int\frac{\mathrm{2}{tdt}}{{t}}=\mathrm{2}{t}+{C}=\mathrm{2}\sqrt{\mathrm{2}{x}−{x}^{\mathrm{2}} −\mathrm{1}}+{C} \\ $$

Answered by hknkrc46 last updated on 01/Apr/23

▶ 2x − x^2 − 1 = −(x − 1)^2 ♥ (√(2x − x^2 − 1)) = (√(−(x − 1)^2 )) = (√(−1)) ∙ (√((x − 1)^2 )) = i∣x − 1∣ = ∓i(1 − x) ▶ ∫ ((2 − 2x)/( (√(2x − x^2 − 1))))dx = ∫ ((2(1 − x))/(∓i(1 − x)))dx = ∓2i∫dx = c ∓ 2ix

$$\blacktriangleright\:\mathrm{2}\boldsymbol{{x}}\:−\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{1}\:=\:−\left(\boldsymbol{{x}}\:−\:\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\:\:\:\heartsuit\:\sqrt{\mathrm{2}\boldsymbol{{x}}\:−\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{1}}\:=\:\sqrt{−\left(\boldsymbol{{x}}\:−\:\mathrm{1}\right)^{\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:=\:\sqrt{−\mathrm{1}}\:\centerdot\:\sqrt{\left(\boldsymbol{{x}}\:−\:\mathrm{1}\right)^{\mathrm{2}} }\:=\:\boldsymbol{{i}}\mid\boldsymbol{{x}}\:−\:\mathrm{1}\mid \\ $$$$\:\:\:\:\:\:\:\:=\:\mp\boldsymbol{{i}}\left(\mathrm{1}\:−\:\boldsymbol{{x}}\right) \\ $$$$\blacktriangleright\:\int\:\frac{\mathrm{2}\:−\:\mathrm{2}\boldsymbol{{x}}}{\:\sqrt{\mathrm{2}\boldsymbol{{x}}\:−\:\boldsymbol{{x}}^{\mathrm{2}} \:−\:\mathrm{1}}}\boldsymbol{{dx}}\: \\ $$$$\:\:\:\:\:=\:\int\:\frac{\mathrm{2}\left(\mathrm{1}\:−\:\boldsymbol{{x}}\right)}{\mp\boldsymbol{{i}}\left(\mathrm{1}\:−\:\boldsymbol{{x}}\right)}\boldsymbol{{dx}}\:=\:\mp\mathrm{2}\boldsymbol{{i}}\int\boldsymbol{{dx}}\: \\ $$$$\:\:\:\:\:=\:\boldsymbol{{c}}\:\mp\:\mathrm{2}\boldsymbol{{ix}} \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply