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Question Number 89286 by M±th+et£s last updated on 16/Apr/20

s h o w t h a t

\int_{0}^{1} \frac{(x^{2} + 1) l n (1 + x)}{x^{4} - x^{2} + 1} d x = \frac{π}{6} l n (2 + \sqrt{3})

Answered by TANMAY PANACEA. last updated on 16/Apr/20

\int \frac{1 + \frac{1}{x^{2}}}{x^{2} + \frac{1}{x^{2}} - 1} . l n (1 + x) d x

l n (1 + x) \int \frac{d {(x - \frac{1}{x})}^{}}{{(x - \frac{1}{x})}^{2} + 1} - \int [\frac{1}{1 + x} \int \frac{d (x - \frac{1}{x})}{{(x - \frac{1}{x})}^{2} + 1} d x] d x

l n (1 + x) . {t a n}^{- 1} (x - \frac{1}{x}) - \int \frac{{t a n}^{- 1} (x - \frac{1}{x})}{1 + x} d x

w a i t \dots

Commented by M±th+et£s last updated on 16/Apr/20

o k s i r

Answered by M±th+et£s last updated on 16/Apr/20