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Question Number 164609 by SANOGO last updated on 19/Jan/22

e n p o s a n t x = t - \frac{1}{t}

{\int_{0}}^{+ o o} \frac{1 + t^{2}}{1 + t^{4}} d t

Answered by Mathspace last updated on 19/Jan/22

I = \int_{0}^{\infty} \frac{1 + \frac{1}{t^{2}}}{t^{2} + \frac{1}{t^{2}}} d t = \int_{0}^{\infty} \frac{1 + \frac{1}{t^{2}}}{{(t - \frac{1}{t})}^{2} + 2} d t

=_{t - \frac{1}{t} = x} \int_{- \infty}^{+ \infty} \frac{d x}{x^{2} + 2}

=_{x = \sqrt{2} y} \int_{- \infty}^{+ \infty} \frac{\sqrt{2} d y}{2 (1 + y^{2})}

= \frac{\sqrt{2}}{2} {[a r c t a n (y)]}_{- \infty}^{+ \infty}

= \frac{\sqrt{2}}{2} (\frac{π}{2} + \frac{π}{2}) = \frac{π \sqrt{2}}{2}

Commented by SANOGO last updated on 19/Jan/22

m e r c i b i e n

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