calculate ∫_0 ^1 ((ln(1+x))/(1+x^2 ))dx


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Question Number 64065 by mathmax by abdo last updated on 12/Jul/19

$c a l c u l a t e \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x$
Commented by mathmax by abdo last updated on 13/Jul/19

$l e t f (t) = \int_{0}^{1} \frac{l n (1 + t x)}{1 + x^{2}} d x w i t h t > 0 w e h a v e$ $f^{^{'}} (t) = \int_{0}^{1} \frac{x}{(1 + t x) (1 + x^{2})} d x l e t d e c o m p o s e$ $F (x) = \frac{x}{(t x + 1) (x^{2} + 1)} \Rightarrow F (x) = \frac{a}{t x + 1} + \frac{b x + c}{x^{2} + 1}$ $a = {l i m}_{x \to - \frac{1}{t}} (t x + 1) F (x) = \frac{- 1}{t (\frac{1}{t^{2}} + 1)} = \frac{- t^{2}}{t (1 + t^{2})} = - \frac{t}{t^{2} + 1}$ ${l i m}_{x \to + \infty} x F (x) = 0 = \frac{a}{t} + b \Rightarrow b = - \frac{a}{t} = \frac{1}{t^{2} + 1} \Rightarrow$ $F (x) = - \frac{t}{(t^{2} + 1) (t x + 1)} + \frac{\frac{x}{t^{2} + 1} + c}{x^{2} + 1}$ $F (0) = 0 = \frac{t}{t^{2} + 1} + c \Rightarrow c = - \frac{t}{t^{2} + 1} \Rightarrow$ $F (x) = \frac{t}{(t^{2} + 1) (t x + 1)} + \frac{1}{t^{2} + 1} \frac{x - t}{x^{2} + 1} \Rightarrow$ $f^{^{'}} (t) = \int_{0}^{1} F (x) d x = \frac{t}{t^{2} + 1} \int_{0}^{1} \frac{d x}{t x + 1} + \frac{1}{t^{2} + 1} \int_{0}^{1} \frac{x - t}{x^{2} + 1} d x$ $= \frac{t}{t^{2} + 1} \frac{1}{t} {[l n (t x + 1)]}_{0}^{1} + \frac{1}{2 (t^{2} + 1)} \int_{0}^{1} \frac{2 x}{x^{2} + 1} d x - \frac{t}{t^{2} + 1} \int_{0}^{1} \frac{d x}{x^{2} + 1}$ $= \frac{l n (t + 1)}{t^{2} + 1} + \frac{1}{2 (t^{2} + 1)} l n (2) - \frac{t}{t^{2} + 1} \frac{π}{4} \Rightarrow$ $f (t) = \int_{0}^{t} \frac{l n (1 + u)}{u^{2} + 1} d u + \frac{l n (2)}{2} \int_{0}^{t} \frac{d u}{1 + u^{2}} - \frac{π}{8} \int_{0}^{t} \frac{2 u}{1 + u^{2}} d u + c$ $= \int_{0}^{t} \frac{l n (1 + u)}{1 + u^{2}} + \frac{l n (2)}{2} a r c t a n (t) - \frac{π}{8} l n (1 + t^{2}) + c = \int_{0}^{1} \frac{l n (1 + t x)}{1 + x^{2}} d x$ $. . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 13/Jul/19

$f (1) = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x = \int_{0}^{1} \frac{l n (1 + u)}{1 + u^{2}} d u + \frac{π}{8} l n (2) - \frac{π}{8} l n (2) + c \Rightarrow$ $c = 0 a n d$ $\int_{0}^{1} \frac{l n (1 + t x)}{1 + x^{2}} d x = \int_{0}^{t} \frac{l n (1 + x)}{1 + x^{2}} d x + \frac{l n (2)}{2} a r c t a n (t) - \frac{π}{8} l n (1 + t^{2})$ $. . . b e c o n t i n u e d . . .$
Commented by mathmax by abdo last updated on 14/Jul/19

$l e t t r y a n o t h e r w a y c h a n g e m e n t x = t a n θ g i v e$ $I = \int_{0}^{1} \frac{l n (1 + x)}{1 + x^{2}} d x = \int_{0}^{\frac{π}{4}} \frac{l n (1 + t a n θ)}{1 + {t a n}^{2} θ} (1 + {t a n}^{2} θ) d θ$ $= \int_{0}^{\frac{π}{4}} l n (1 + t a n θ) d θ c h a n g θ = \frac{π}{4} - u \Rightarrow$ $\int_{0}^{\frac{π}{4}} l n (1 + t a n θ) d θ = \int_{\frac{π}{4}}^{0} l n (1 + \frac{1 - t a n u}{1 + t a n u}) (- d u)$ $= \int_{0}^{\frac{π}{4}} l n (\frac{2}{1 + t a n u}) d u = \frac{π}{4} l n (2) - \int_{0}^{\frac{π}{4}} l n (1 + t a n u) d u$ $\Rightarrow I = \frac{π}{4} l n (2) - I \Rightarrow 2 I = \frac{π}{4} l n (2) \Rightarrow I = \frac{π}{8} l n (2)$
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