0-1-ln-1-x-x-2-1-dx-

Question Number 86646 by Ar Brandon last updated on 30/Mar/20

\int_{0}^{1} \frac{\ln (1 + x)}{x^{2} + 1} dx

Commented by mathmax by abdo last updated on 30/Mar/20

I = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x c h a 7 g e m e n t x = t a n t g i v e

I = \int_{0}^{\frac{π}{4}} \frac{l n (1 + t a n (t))}{1 + {t a n}^{2} t} (1 + {t a n}^{2} t) d t

= \int_{0}^{\frac{π}{4}} l n (1 + t a n t) d t

=_{t = \frac{π}{4} - u} \int_{0}^{\frac{π}{4}} l n (1 + t a n (\frac{π}{4} - u)) d u

= \int_{0}^{\frac{π}{4}} l n (1 + \frac{1 - t a n u}{1 + t a n u}) d u = \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n u}) d u

= \frac{π}{4} l n (2) - \int_{0}^{\frac{π}{4}} l n (1 + t a n u) d u \Rightarrow

2 I = \frac{π}{4} l n (2) \Rightarrow I = \frac{π}{8} l n (2)

Commented by Ar Brandon last updated on 30/Mar/20

c o o l!!