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Question Number 108034 by mathdave last updated on 14/Aug/20

	Answered by mathmax by abdo last updated on 14/Aug/20
	$I =∫_(49) ^(196) (dx/(((√x)−2)(√(2+(√x))))) we do tbe cha7gement (√x)=t ⇒ I =∫_7 ^(14) ((2t dt)/((t−2)(√(2+t)))) dt =2 ∫_7 ^(14) ((tdt)/((t−2)(√(2+t)))) =_((√(2+t))=u) 2∫_3 ^4 ((u^2 −2)/((u^2 −4)u)) (2u)du =4 ∫_3 ^4 ((u^2 −2)/(u^2 −4))du =4 ∫_3 ^4 ((u^2 −4+2)/(u^2 −4))du =4 ∫_3 ^4 du +8 ∫_3 ^4 (du/((u−2)(u+2))) =4 +2∫_3 ^4 ((1/(u−2))−(1/(u+2)))du =4 +2 [ln∣((u−2)/(u+2))∣]_3 ^4 =4 +2{ln((2/6))−ln((1/5))} =4+2{−ln(3)+ln)5)} =4 +2ln((5/3)) =ln(e^4 )+ln(((25)/9)) =ln(((25)/9)e^4 )$
	$I = \int_{49}^{196} \frac{dx}{(\sqrt{x} - 2) \sqrt{2 + \sqrt{x}}} we do tbe cha 7 gement \sqrt{x} = t \Rightarrow$ $I = \int_{7}^{14} \frac{2 t dt}{(t - 2) \sqrt{2 + t}} dt = 2 \int_{7}^{14} \frac{tdt}{(t - 2) \sqrt{2 + t}}$ $=_{\sqrt{2 + t} = u} 2 \int_{3}^{4} \frac{u^{2} - 2}{(u^{2} - 4) u} (2 u) du = 4 \int_{3}^{4} \frac{u^{2} - 2}{u^{2} - 4} du$ $= 4 \int_{3}^{4} \frac{u^{2} - 4 + 2}{u^{2} - 4} du = 4 \int_{3}^{4} du + 8 \int_{3}^{4} \frac{du}{(u - 2) (u + 2)}$ $= 4 + 2 \int_{3}^{4} (\frac{1}{u - 2} - \frac{1}{u + 2}) du$ $= 4 + 2 {[\ln ∣ \frac{u - 2}{u + 2} ∣]}_{3}^{4} = 4 + 2 {\ln (\frac{2}{6}) - \ln (\frac{1}{5})}$ $= 4 + 2 {- \ln (3) + \ln) 5)} = 4 + 2 \ln (\frac{5}{3}) = \ln (e^{4}) + \ln (\frac{25}{9})$ $= \ln (\frac{25}{9} e^{4})$
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