M= ∫ _(−15) ^(−8) ( (dx/(x(√(1−x))))) ?


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Question Number 121119 by benjo_mathlover last updated on 05/Nov/20

$M = \int \underset{- 15}{\overset{- 8}{}} (\frac{dx}{x \sqrt{1 - x}}) ?$
	Answered by liberty last updated on 05/Nov/20
	$M =∫_(−15) ^(−8) (dx/(x^2 (√(x^(−1) −1)))) let u = x^(−1) −1 ⇒du=−x^(−2) dx → { ((u=−(9/8))),((u=−((16)/(15)))) :} M=−∫_(−16/15) ^(−9/8) (du/u^(1/2) ) = −2(√u) ]_(−16/15) ^(−9/8) = −((3i)/( (√2))) −(−((8i)/( (√(15)))))=((8/( (√(15))))−(3/( (√2))))i$
	$M = \underset{- 15}{\int^{- 8}} \frac{dx}{x^{2} \sqrt{x^{- 1} - 1}}$ $let u = x^{- 1} - 1 \Rightarrow du = - x^{- 2} dx \to {\begin{cases} u = - \frac{9}{8} \\ u = - \frac{16}{15} \end{cases}$ ${M = - \underset{- 16 / 15}{\int^{- 9 / 8}} \frac{du}{u^{1 / 2}} = - 2 \sqrt{u}]}_{- 16 / 15}^{- 9 / 8}$ $= - \frac{3 i}{\sqrt{2}} - (- \frac{8 i}{\sqrt{15}}) = (\frac{8}{\sqrt{15}} - \frac{3}{\sqrt{2}}) i$
	Answered by 675480065 last updated on 05/Nov/20
	$u^2 =1−x ⇒ du=−dx {x=1−u^2 } U=∫_u_1 ^( u_2 ) (du/((u^2 −1)u)) =∫_u_1 ^( u_2 ) (du/((u−1)(u+1)u)) M=(1/2)ln∣(√(1−x))+1∣+(1/2)ln∣(√(1−x))−1∣−ln∣(√(1−x))∣]_(−15) ^(−8) M=[ln∣(x^2 /( (√(1−x))))∣]_(−15) ^(−8) =ln(((64)/3)×(4/(225)))$
	$u^{2} = 1 - x \Rightarrow du = - dx {x = 1 - u^{2}}$ $U = \int_{u_{1}}^{u_{2}} \frac{du}{(u^{2} - 1) u} = \int_{u_{1}}^{u_{2}} \frac{du}{(u - 1) (u + 1) u}$ ${M = \frac{1}{2} \ln ∣ \sqrt{1 - x} + 1 ∣ + \frac{1}{2} \ln ∣ \sqrt{1 - x} - 1 ∣ - \ln ∣ \sqrt{1 - x} ∣]}_{- 15}^{- 8}$ $M = {[\ln ∣ \frac{x^{2}}{\sqrt{1 - x}} ∣]}_{- 15}^{- 8} = \ln (\frac{64}{3} \times \frac{4}{225})$
	Answered by MJS_new last updated on 05/Nov/20

	$\int \frac{d x}{x \sqrt{1 - x}} =$ $[t = \frac{1 + \sqrt{1 - x}}{\sqrt{x}} \to d x = - 2 (x - 1 + \sqrt{1 - x}) \sqrt{x} d t]$ $= - 2 \int \frac{d t}{t} = - 2 \ln ∣ t ∣ \Rightarrow M = \ln \frac{5}{6}$
	Answered by Bird last updated on 05/Nov/20

	$M = \int_{- 15}^{- 8} \frac{d x}{x \sqrt{1 - x}} c h a n g e m e n t$ $\sqrt{1 - x} = t g i v e 1 - x = t^{2} \Rightarrow x = 1 - t^{2}$ $\Rightarrow M = \int_{4}^{3} \frac{- 2 t d t}{(1 - t^{2}) t} = 2 \int_{3}^{4} \frac{d t}{1 - t^{2}}$ $= \int_{3}^{4} (\frac{1}{1 - t} + \frac{1}{1 + t}) d t$ $= {[l n ∣ \frac{1 + t}{1 - t} ∣]}_{3}^{4} = l n (\frac{5}{3}) - l n (\frac{4}{2})$ $= l n 5 - l n 3 - l n 2 = l n 5 - l n 6$
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