find the value of ∫_0 ^1 (x^2 /(1+x^4 ))dx


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Question Number 34228 by abdo imad last updated on 03/May/18

$f i n d t h e v a l u e o f \int_{0}^{1} \frac{x^{2}}{1 + x^{4}} d x$
Commented by tanmay.chaudhury50@gmail.com last updated on 03/May/18
$let I=∫(x^2 /(1+x^(4 ) ))dx =∫(1/(x^2 +(1/x^2 )))dx =(1/2)∫(((1+(1/x^2 ))+(1−(1/x^2 )))/(x^2 +(1/x^2 )))dx =(1/2)∫(((1+(1/x^2 )))/((x−(1/x))^2 +2))+(1/2)∫(((1−(1/x^2 )))/((x+(1/x))^2 −2)) =(1/2)∫((d(x−(1/x)))/((x−(1/x))^2 +2))+(1/2)∫((d(x+(1/x)))/((x+(1/x))^2 −2)) =(1/2).(1/(√2)).tan^(−1) {(((x−(1/x)))/(√2))}+(1/2).(1/(2(√2))).ln∣(((√2) −(x+(1/x)))/((√2)+(x+(1/x))) now given intregal has upper limit 1 and lower limit limit 0 put the limit$
$l e t I = \int \frac{x^{2}}{1 + x^{4}} d x$ $= \int \frac{1}{x^{2} + \frac{1}{x^{2}}} d x$ $= \frac{1}{2} \int \frac{(1 + \frac{1}{x^{2}}) + (1 - \frac{1}{x^{2}})}{x^{2} + \frac{1}{x^{2}}} d x$ $= \frac{1}{2} \int \frac{(1 + \frac{1}{x^{2}})}{{(x - \frac{1}{x})}^{2} + 2} + \frac{1}{2} \int \frac{(1 - \frac{1}{x^{2}})}{{(x + \frac{1}{x})}^{2} - 2}$ $= \frac{1}{2} \int \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 2} + \frac{1}{2} \int \frac{d (x + \frac{1}{x})}{{(x + \frac{1}{x})}^{2} - 2}$ $= \frac{1}{2} . \frac{1}{\sqrt{2}} . {t a n}^{- 1} {\frac{(x - \frac{1}{x})}{\sqrt{2}}} + \frac{1}{2} . \frac{1}{2 \sqrt{2}} . l n ∣ \frac{\sqrt{2} - (x + \frac{1}{x})}{\sqrt{2} + (x + \frac{1}{x}}$ $n o w g i v e n i n t r e g a l h a s u p p e r l i m i t 1 a n d l o w e r l i m i t$ $l i m i t 0$ $p u t t h e l i m i t$
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