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1-8-x-2-4-x-2-dx-2-x-6-2x-2-5x-3-




Question Number 136968 by Eric002 last updated on 28/Mar/21
1)∫(8/(x^2 (√(4−x^2 ))))dx  2)∫((x−6 )/(2x^2 −5x+3))
$$\left.\mathrm{1}\right)\int\frac{\mathrm{8}}{{x}^{\mathrm{2}} \sqrt{\mathrm{4}−{x}^{\mathrm{2}} }}{dx} \\ $$$$\left.\mathrm{2}\right)\int\frac{{x}−\mathrm{6}\:}{\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{3}} \\ $$
Answered by EDWIN88 last updated on 28/Mar/21
(1)∫ (8/(x^3  (√(4x^(−2) −1)))) dx = ∫ ((8x^(−3) )/( (√(4x^(−2) −1)))) dx  set u^2  = 4x^(−2) −1   2u du =−8x^(−3)  dx  E = −2∫ (u/u) du =−2u + c   E=−2 (√(4x^(−2) −1)) + c = −2(√((4−x^2 )/x^2 )) + c  E = −((2(√(4−x^2 )))/x) + c
$$\left(\mathrm{1}\right)\int\:\frac{\mathrm{8}}{\mathrm{x}^{\mathrm{3}} \:\sqrt{\mathrm{4x}^{−\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx}\:=\:\int\:\frac{\mathrm{8x}^{−\mathrm{3}} }{\:\sqrt{\mathrm{4x}^{−\mathrm{2}} −\mathrm{1}}}\:\mathrm{dx} \\ $$$$\mathrm{set}\:\mathrm{u}^{\mathrm{2}} \:=\:\mathrm{4x}^{−\mathrm{2}} −\mathrm{1}\: \:\mathrm{2u}\:\mathrm{du}\:=−\mathrm{8x}^{−\mathrm{3}} \:\mathrm{dx} \\ $$$$\mathrm{E}\:=\:−\mathrm{2}\int\:\frac{\mathrm{u}}{\mathrm{u}}\:\mathrm{du}\:=−\mathrm{2u}\:+\:\mathrm{c}\: \\ $$$$\mathrm{E}=−\mathrm{2}\:\sqrt{\mathrm{4x}^{−\mathrm{2}} −\mathrm{1}}\:+\:\mathrm{c}\:=\:−\mathrm{2}\sqrt{\frac{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }}\:+\:\mathrm{c} \\ $$$$\mathrm{E}\:=\:−\frac{\mathrm{2}\sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}{\mathrm{x}}\:+\:\mathrm{c}\: \\ $$
Answered by EDWIN88 last updated on 28/Mar/21
(2)E=∫ ((x−6)/((2x+1)(x−3))) dx   Partial fraction  ((x−6)/(2x^2 −5x+3)) = (p/(2x+1)) + (q/(x−3))  where  { ((p=[((x−6)/(x−3)) ] _(x=−(1/2)) = ((−13)/(−7))=((13)/7))),((q=[((x−6)/(2x+1)) ]_(x=3) =−(3/7))) :}  E=((13)/7)∫ (dx/(2x+1)) −(3/7)∫ (dx/(x−3))  E=((13)/(14)) ln ∣2x+1∣−(3/7) ln ∣x−3∣ + c
$$\left(\mathrm{2}\right)\mathrm{E}=\int\:\frac{\mathrm{x}−\mathrm{6}}{\left(\mathrm{2x}+\mathrm{1}\right)\left(\mathrm{x}−\mathrm{3}\right)}\:\mathrm{dx}\: \\ $$$$\mathrm{Partial}\:\mathrm{fraction}\: \frac{\mathrm{x}−\mathrm{6}}{\mathrm{2x}^{\mathrm{2}} −\mathrm{5x}+\mathrm{3}}\:=\:\frac{\mathrm{p}}{\mathrm{2x}+\mathrm{1}}\:+\:\frac{\mathrm{q}}{\mathrm{x}−\mathrm{3}} \\ $$$$\mathrm{where}\:\begin{cases}{\mathrm{p}=\left[\frac{\mathrm{x}−\mathrm{6}}{\mathrm{x}−\mathrm{3}}\:\right]\:_{\mathrm{x}=−\frac{\mathrm{1}}{\mathrm{2}}} =\:\frac{−\mathrm{13}}{−\mathrm{7}}=\frac{\mathrm{13}}{\mathrm{7}}}\\{\mathrm{q}=\left[\frac{\mathrm{x}−\mathrm{6}}{\mathrm{2x}+\mathrm{1}}\:\right]_{\mathrm{x}=\mathrm{3}} =−\frac{\mathrm{3}}{\mathrm{7}}}\end{cases} \\ $$$$\mathrm{E}=\frac{\mathrm{13}}{\mathrm{7}}\int\:\frac{\mathrm{dx}}{\mathrm{2x}+\mathrm{1}}\:−\frac{\mathrm{3}}{\mathrm{7}}\int\:\frac{\mathrm{dx}}{\mathrm{x}−\mathrm{3}} \\ $$$$\mathrm{E}=\frac{\mathrm{13}}{\mathrm{14}}\:\mathrm{ln}\:\mid\mathrm{2x}+\mathrm{1}\mid−\frac{\mathrm{3}}{\mathrm{7}}\:\mathrm{ln}\:\mid\mathrm{x}−\mathrm{3}\mid\:+\:\mathrm{c}\: \\ $$
Answered by mathmax by abdo last updated on 28/Mar/21
1) I=∫ (8/(x^2 (√(4−x^2 ))))dx ⇒I=_(x=2sinθ)   ∫  (8/(4sin^2 θ.2cosθ))(2cosθ)dθ  =4∫ (dθ/(sin^2 θ)) =8 ∫ (dθ/(1−cos(2θ)))=_(2θ=t)    8∫  (dt/(2(1−cost)))=4∫ (dt/(1−cost))  =_(tan((t/2))=y)    4 ∫  ((2dy)/((1+y^2 )(1−((1−y^2 )/(1+y^2 )))))=8∫  (dy/(1+y^2 −1+y^2 )) =4∫ (dy/y^2 )  =−(4/y)+C =−(4/(tan((t/2))))+C =−(4/(tan(θ)))+C=−(4/(tan(arcsin((x/2)))))+C
$$\left.\mathrm{1}\right)\:\mathrm{I}=\int\:\frac{\mathrm{8}}{\mathrm{x}^{\mathrm{2}} \sqrt{\mathrm{4}−\mathrm{x}^{\mathrm{2}} }}\mathrm{dx}\:\Rightarrow\mathrm{I}=_{\mathrm{x}=\mathrm{2sin}\theta} \:\:\int\:\:\frac{\mathrm{8}}{\mathrm{4sin}^{\mathrm{2}} \theta.\mathrm{2cos}\theta}\left(\mathrm{2cos}\theta\right)\mathrm{d}\theta \\ $$$$=\mathrm{4}\int\:\frac{\mathrm{d}\theta}{\mathrm{sin}^{\mathrm{2}} \theta}\:=\mathrm{8}\:\int\:\frac{\mathrm{d}\theta}{\mathrm{1}−\mathrm{cos}\left(\mathrm{2}\theta\right)}=_{\mathrm{2}\theta=\mathrm{t}} \:\:\:\mathrm{8}\int\:\:\frac{\mathrm{dt}}{\mathrm{2}\left(\mathrm{1}−\mathrm{cost}\right)}=\mathrm{4}\int\:\frac{\mathrm{dt}}{\mathrm{1}−\mathrm{cost}} \\ $$$$=_{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)=\mathrm{y}} \:\:\:\mathrm{4}\:\int\:\:\frac{\mathrm{2dy}}{\left(\mathrm{1}+\mathrm{y}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{1}−\mathrm{y}^{\mathrm{2}} }{\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\right)}=\mathrm{8}\int\:\:\frac{\mathrm{dy}}{\mathrm{1}+\mathrm{y}^{\mathrm{2}} −\mathrm{1}+\mathrm{y}^{\mathrm{2}} }\:=\mathrm{4}\int\:\frac{\mathrm{dy}}{\mathrm{y}^{\mathrm{2}} } \\ $$$$=−\frac{\mathrm{4}}{\mathrm{y}}+\mathrm{C}\:=−\frac{\mathrm{4}}{\mathrm{tan}\left(\frac{\mathrm{t}}{\mathrm{2}}\right)}+\mathrm{C}\:=−\frac{\mathrm{4}}{\mathrm{tan}\left(\theta\right)}+\mathrm{C}=−\frac{\mathrm{4}}{\mathrm{tan}\left(\mathrm{arcsin}\left(\frac{\mathrm{x}}{\mathrm{2}}\right)\right)}+\mathrm{C} \\ $$

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