calculate ∫_0 ^1 ((√x)/(1+x^2 ))dx .


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Question Number 38199 by prof Abdo imad last updated on 22/Jun/18

$c a l c u l a t e \int_{0}^{1} \frac{\sqrt{x}}{1 + x^{2}} d x .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jun/18
	$x=t^(2 ) dx=2tdt ∫_0 ^1 ((2t^2 )/(1+t^4 ))dt ∫_0 ^1 (2/(t^2 +(1/t^2 )))dt ∫_0 ^1 ((1−(1/t^2 ))/((t+(1/t))^2 −2))+∫_0 ^1 ((1+(1/t^2 ))/((t−(1/t))^2 +2)) ∫_0 ^1 ((d(t+(1/t)))/((t+(1/t))^2 −2))+∫_0 ^1 ((d(t−(1/t)))/((t−(1/t))^2 +2)) (1/(2(√2))){ln∣(((t+(1/t))−(√2))/((t+(1/t))+(√2)))∣+(1/(√2))tan^(−1) (((t−(1/t))/(√2)))}_0 ^1 =$
	$x = t^{2} d x = 2 t d t$ $\int_{0}^{1} \frac{2 t^{2}}{1 + t^{4}} d t$ $\int_{0}^{1} \frac{2}{t^{2} + \frac{1}{t^{2}}} d t$ $\int_{0}^{1} \frac{1 - \frac{1}{t^{2}}}{{(t + \frac{1}{t})}^{2} - 2} + \int_{0}^{1} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2}$ $\int_{0}^{1} \frac{d (t + \frac{1}{t})}{{(t + \frac{1}{t})}^{2} - 2} + \int_{0}^{1} \frac{d (t - \frac{1}{t})}{{(t - \frac{1}{t})}^{2} + 2}$ $\frac{1}{2 \sqrt{2}} {l n ∣ \frac{(t + \frac{1}{t}) - \sqrt{2}}{(t + \frac{1}{t}) + \sqrt{2}} ∣ + \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{t - \frac{1}{t}}{\sqrt{2}})}_{0}^{1}$ $=$
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