∫((2x^2 −1+2x(√(x^2 −1)))/(x^2 −x+(x−1)(√(x^2 −1))))dx=? ∫(dx/(x(√(x+1))(√((1−x)^3 ))))=?


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Question Number 73202 by MJS last updated on 08/Nov/19

$\int \frac{2 x^{2} - 1 + 2 x \sqrt{x^{2} - 1}}{x^{2} - x + (x - 1) \sqrt{x^{2} - 1}} d x = ?$ $\int \frac{d x}{x \sqrt{x + 1} \sqrt{{(1 - x)}^{3}}} = ?$
Commented by mathmax by abdo last updated on 09/Nov/19

$l e t A = \int \frac{d x}{x \sqrt{x + 1} {(1 - x)}^{\frac{3}{2}}} c h a n g e m e n t x = c o s 2 t g i v e$ $A = \int \frac{- 2 s i n (2 t) d t}{c o s (2 t) \sqrt{2 {c o s}^{2} t} {(2 {s i n}^{2} t)}^{\frac{3}{2}}} = - 2 \int \frac{s i n (2 t) d t}{c o s (2 t) \sqrt{2} c o s t 2^{\frac{3}{2}} {s i n}^{3} t}$ $= - 2 \int \frac{2 s i n t c o s t}{4 c o s (2 t) c o s t s i n t {s i n}^{2} t} d t$ $= - \int \frac{d t}{c o s (2 t) (\frac{1 - c o s (2 t)}{2})} = - 2 \int \frac{d t}{c o s (2 t) - {c o s}^{2} (2 t)}$ $= - 2 \int \frac{d t}{c o s (2 t) - \frac{1 + c o s (4 t)}{2}} = - 4 \int \frac{d t}{2 c o s (2 t) - c o s (4 t) - 1}$ $=_{2 t = z} - 4 \int \frac{d z}{2 (2 c o s (z) - c o s (2 z) - 1)}$ $= - 2 \int \frac{d z}{2 c o s z - c o s (2 z) - 1}$ $= - 2 \int \frac{d z}{2 c o s (z) - (2 {c o s}^{2} z - 1) - 1}$ $= - 2 \int \frac{d z}{- 2 {c o s}^{2} z + 2 c o s z} = \int \frac{d z}{{c o s}^{2} z - c o s z}$ $= \int \frac{d z}{c o s z (c o s z - 1)} = \int (\frac{1}{c o s z - 1} - \frac{1}{c o s z}) d z$ $= \int \frac{d z}{c o s z - 1} - \int \frac{d z}{c o s z} w e h a v e$ $\int \frac{d z}{c o s z - 1} =_{t a n (\frac{z}{2}) = u} \int \frac{1}{\frac{1 - u^{2}}{1 + u^{2}} - 1} \frac{2 d u}{1 + u^{2}}$ $= \int \frac{2 d u}{1 - u^{2} - 1 - u^{2}} = \int - \frac{d u}{u^{2}} = \frac{1}{u} + C = \frac{1}{t a n (\frac{z}{2})} + c$ $= \frac{1}{t a n (t)} + C = \frac{1}{t a n (\frac{a r c o s x}{2})} + C a l s o$ $\int \frac{d z}{c o s z} =_{t a n (\frac{z}{2}) = u} \int \frac{1}{\frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}} = \int \frac{2 d u}{1 - u^{2}} = \int (\frac{1}{1 + u} + \frac{1}{1 - u}) d u$ $= l n ∣ \frac{1 + u}{1 - u} ∣ + c = l n ∣ \frac{1 + t a n (\frac{z}{2})}{1 - t a n (\frac{z}{2})} ∣ + c = l n ∣ \frac{1 + t a n (t)}{1 - t a n (t)} ∣ + c$ $= l n ∣ \frac{1 + t a n (\frac{a r c o s x}{2})}{1 - t a n (\frac{a r c o s x}{2})} ∣ + c \Rightarrow$ $A = \frac{1}{t a n (\frac{a r c o s x}{2})} - l n ∣ \frac{1 + t a n (\frac{a r c o s x}{2})}{1 - t a n (\frac{a r c o s x}{2})} ∣ + C .$
Commented by MJS last updated on 10/Nov/19

$thank you$
Commented by mathmax by abdo last updated on 10/Nov/19

$y o u a r e w e l c o m e .$
	Answered by MJS last updated on 10/Nov/19

	$\int \frac{d x}{x \sqrt{x + 1} \sqrt{{(1 - x)}^{3}}} =$ $[t = \frac{\sqrt{1 + x}}{\sqrt{1 - x}} \to d x = \sqrt{x + 1} \sqrt{{(1 - x)}^{3}} d t]$ $= \int \frac{t^{2} + 1}{t^{2} - 1} d t =$ $= \int d t + \int \frac{d t}{t - 1} - \int \frac{d t}{t + 1} =$ $= t + \ln \frac{t - 1}{t + 1} = \frac{\sqrt{1 + x}}{\sqrt{1 - x}} + \ln \frac{1 - \sqrt{1 - x^{2}}}{x} + C$
	Answered by MJS last updated on 10/Nov/19

	$\int \frac{2 x^{2} - 1 + 2 x \sqrt{x^{2} - 1}}{x^{2} - x + (x - 1) \sqrt{x^{2} - 1}} d x =$ $[t = x + \sqrt{x^{2} - 1} \to d x = \frac{\sqrt{x^{2} - 1}}{x + \sqrt{x^{2} - 1}} d t]$ $= \int \frac{t + 1}{t - 1} d t = t + 2 \ln (t - 1) =$ $= x + \sqrt{x^{2} - 1} + 2 \ln ∣ x - 1 + \sqrt{x^{2} - 1} ∣ + C$
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