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Question Number 45706 by Meritguide1234 last updated on 15/Oct/18

	Answered by tanmay.chaudhury50@gmail.com last updated on 17/Oct/18
	$trying to solve... ∫(((1+x^4 ))/((1−x^4 )(√(1+x^4 )) ))dx ∫((1+x^4 )/(x^2 ((1/x^2 )−x^2 )×x(√((1/x^2 )+x^2 )) ))dx ∫(((1/x^3 )+x)/(((1/x^2 )−x^2 )×{((1/x^2 )+x^2 )^2 }^(1/4) ))dx ∫((((1/x^3 )+x)dx)/(((1/x^2 )−x^2 )×{((1/x^2 )−x^2 )^2 +4}^(1/4) )) t=(1/x^2 )−x^2 (dt/dx)=((−2)/x^3 )−2x (dt/(−2))=((1/x^3 )+x)dx ∫(dt/(−2×t×(t^2 +4)^(1/4) )) ((−1)/2)∫((tdt)/(t^2 ×(t^2 +4)^(1/4) )) y^4 =t^2 +4 4y^3 ×(dy/dt)=2t ((−1)/2)∫((2y^3 dy)/((y^4 −4)×y)) =−1∫((y^2 dy)/((y^2 +1)(y^2 −1))) =((−1)/2)∫((y^2 +1+y^2 −1)/((y^2 +1)(y^2 −1)))dy ((−1)/2)[∫(dy/(y^2 −1))+∫(dy/(y^2 +1))] =((−1)/2)[{(1/2)∫(((y+1)−(y−1})/((y+1)(y−1)))+∫(dy/(y^2 +1))] ((−1)/2)[{(1/2)×∫(dy/(y−1))−(1/2)∫(dy/(y+1))+∫(dy/(y^2 +1)) dy] =((−1)/2)[(1/2)ln(((y−1)/(y+1)))+tan^(−1) y] =((−1)/2)[(1/2)ln{(((t^2 +4)^(1/4) −1)/((t^2 +4)^(1/4) +1))}+tan^(−1) (t^2 +4)^(1/4) ]+c =((−1)/2)[(1/2)ln∣(({((1/x^2 )−x^2 )^2 +4}^(1/4) −1)/({((1/x^2 )−x^2 )^2 +4}^(1/4) +2))∣+tan^(−1) {((1/x^2 )−x)^2 +4}^(1/4) +c$
	$t r y i n g t o s o l v e . . .$ $\int \frac{(1 + x^{4})}{(1 - x^{4}) \sqrt{1 + x^{4}}} d x$ $\int \frac{1 + x^{4}}{x^{2} (\frac{1}{x^{2}} - x^{2}) \times x \sqrt{\frac{1}{x^{2}} + x^{2}}} d x$ $\int \frac{\frac{1}{x^{3}} + x}{(\frac{1}{x^{2}} - x^{2}) \times {{(\frac{1}{x^{2}} + x^{2})}^{2}}^{\frac{1}{4}}} d x$ $\int \frac{(\frac{1}{x^{3}} + x) d x}{(\frac{1}{x^{2}} - x^{2}) \times {{(\frac{1}{x^{2}} - x^{2})}^{2} + 4}^{\frac{1}{4}}}$ $t = \frac{1}{x^{2}} - x^{2} \frac{d t}{d x} = \frac{- 2}{x^{3}} - 2 x \frac{d t}{- 2} = (\frac{1}{x^{3}} + x) d x$ $\int \frac{d t}{- 2 \times t \times {(t^{2} + 4)}^{\frac{1}{4}}}$ $\frac{- 1}{2} \int \frac{t d t}{t^{2} \times {(t^{2} + 4)}^{\frac{1}{4}}}$ $y^{4} = t^{2} + 4 4 y^{3} \times \frac{d y}{d t} = 2 t$ $\frac{- 1}{2} \int \frac{2 y^{3} d y}{(y^{4} - 4) \times y}$ $= - 1 \int \frac{y^{2} d y}{(y^{2} + 1) (y^{2} - 1)}$ $= \frac{- 1}{2} \int \frac{y^{2} + 1 + y^{2} - 1}{(y^{2} + 1) (y^{2} - 1)} d y$ $\frac{- 1}{2} [\int \frac{d y}{y^{2} - 1} + \int \frac{d y}{y^{2} + 1}]$ $= \frac{- 1}{2} [{\frac{1}{2} \int \frac{(y + 1) - (y - 1}}{(y + 1) (y - 1)} + \int \frac{d y}{y^{2} + 1}]$ $\frac{- 1}{2} [{\frac{1}{2} \times \int \frac{d y}{y - 1} - \frac{1}{2} \int \frac{d y}{y + 1} + \int \frac{d y}{y^{2} + 1} d y]$ $= \frac{- 1}{2} [\frac{1}{2} l n (\frac{y - 1}{y + 1}) + {t a n}^{- 1} y]$ $= \frac{- 1}{2} [\frac{1}{2} l n {\frac{{(t^{2} + 4)}^{\frac{1}{4}} - 1}{{(t^{2} + 4)}^{\frac{1}{4}} + 1}} + {t a n}^{- 1} {(t^{2} + 4)}^{\frac{1}{4}}] + c$ $= \frac{- 1}{2} [\frac{1}{2} l n ∣ \frac{{{(\frac{1}{x^{2}} - x^{2})}^{2} + 4}^{\frac{1}{4}} - 1}{{{(\frac{1}{x^{2}} - x^{2})}^{2} + 4}^{\frac{1}{4}} + 2} ∣ + {t a n}^{- 1} {{(\frac{1}{x^{2}} - x)}^{2} + 4}^{\frac{1}{4}} + c$
	Commented by Meritguide1234 last updated on 18/Oct/18

	$n i c e$
	Commented by tanmay.chaudhury50@gmail.com last updated on 18/Oct/18

	$t h a n k y o u s i r . . .$
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