calculate-0-1-1-x-3-1-x-4-dx-

Question Number 47058 by maxmathsup by imad last updated on 04/Nov/18

c a l c u l a t e \int_{0}^{1} \frac{1 + x^{3}}{1 + x^{4}} d x

Answered by tanmay.chaudhury50@gmail.com last updated on 04/Nov/18

\int \frac{d x}{1 + x^{4}} + \frac{1}{4} \int \frac{4 x^{3}}{1 + x^{4}} d x

I_{1} + I_{2}

I_{2} = \frac{1}{4} \int \frac{4 x^{3}}{1 + x^{4}} d x

= \frac{1}{4} l n (1 + x^{4}) + c_{2}

j_{2} =∣ I_{2} ∣_{0}^{1} = \frac{1}{4} l n (1 + 1) - \frac{1}{4} l n (1 + 0) \to \frac{1}{4} l n 2

I_{1} = \frac{1}{2} \int \frac{\frac{2}{x^{2}}}{\frac{1}{x^{2}} + 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{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 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 \sqrt{2}} l n ∣ (\frac{x + \frac{1}{x} - \sqrt{2}}{x + \frac{1}{x} + \sqrt{2}}) ∣] + c_{1}

= \frac{1}{2} [\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{x^{2} - 1}{x \sqrt{2}}) - \frac{1}{2 \sqrt{2}} l n ∣ (\frac{x^{2} + 1 - x \sqrt{2}}{x^{2} + 1 + x \sqrt{2}}) ∣

j_{1} =∣ I_{1} ∣_{0}^{1}

j_{1} = \frac{1}{2} [{\frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{1 - 1}{1 \times \sqrt{2}}) - \frac{1}{\sqrt{2}} {t a n}^{- 1} (\frac{0 - 1}{0 \times \sqrt{2}})} -

\frac{1}{2 \sqrt{2}} l n ∣ (\frac{1 + 1 - \sqrt{2}}{1 + 1 + \sqrt{2}}) ∣ - \frac{1}{2 \sqrt{2}} l n ∣ (\frac{1}{1})]

= \frac{1}{2} [{\frac{1}{\sqrt{2}} \times 0 - \frac{1}{\sqrt{2}} (- \frac{π}{2})} - \frac{1}{2 \sqrt{2}} l n (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})]

= \frac{π}{4 \sqrt{2}} - \frac{1}{2 \sqrt{2}} l n (\frac{2 - \sqrt{2}}{2 + \sqrt{2}})

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