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Question Number 138485 by SLVR last updated on 14/Apr/21

Commented by SLVR last updated on 14/Apr/21

$G o o d m o r n i n g m r . W t h a n k s f o r$ $y o u r s u p p o r t . T h e a b o v e i s a l s o i$ $m i s s e d l o n g a g o . . a n d i c o u l d n o t$ $r e t r i v e f r o m t h e g r o u p . k i n d l y h e l p$
	Answered by phanphuoc last updated on 14/Apr/21

	$u = a r c x s i n x d v = 1 / \sqrt{1 - x + x^{2}} d x$
	Answered by Ñï= last updated on 14/Apr/21
	$∫_0 ^1 ((sin^(−1) (√x))/( (√(1−x+x^2 ))))dx=∫_0 ^1 ((sin^(−1) (√(1−x)))/( (√(1−x+x^2 ))))dx=(1/2)∙(π/2)∫_0 ^1 (dx/( (√((x−(1/2))^2 +(3/4))))) =(π/4)ln(x−(1/2)+(√(1−x+x^2 )))_0 ^1 =(π/4)ln 3 {sin^(−1) (√x)+sin^(−1) (√(1−x))=(π/2),∫(du/( (√(u^2 +a^2 ))))=ln∣u+(√(u^2 +a^2 ))∣+C} ∫_0 ^1 ((sin^(−1) x)/( (√(1−x+x^2 ))))dx=(π/4)ln 3$
	$\int_{0}^{1} \frac{si n^{- 1} \sqrt{x}}{\sqrt{1 - x + x^{2}}} d x = \int_{0}^{1} \frac{\sin^{- 1} \sqrt{1 - x}}{\sqrt{1 - x + x^{2}}} d x = \frac{1}{2} \cdot \frac{π}{2} \int_{0}^{1} \frac{d x}{\sqrt{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}}$ $= \frac{π}{4} l n {(x - \frac{1}{2} + \sqrt{1 - x + x^{2}})}_{0}^{1} = \frac{π}{4} l n 3$ ${\sin^{- 1} \sqrt{x} + \sin^{- 1} \sqrt{1 - x} = \frac{π}{2}, \int \frac{d u}{\sqrt{u^{2} + a^{2}}} = l n ∣ u + \sqrt{u^{2} + a^{2}} ∣ + C}$ $\int_{0}^{1} \frac{\sin^{- 1} x}{\sqrt{1 - x + x^{2}}} d x = \frac{π}{4} l n 3$
	Commented by SLVR last updated on 14/Apr/21

	$t h a n k s s i r . . . i a m w r o n g w r i t i n g$ $t h e q u e s t i o n .$
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