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

Question Number 92087 by mathmax by abdo last updated on 04/May/20

c a l c u l a t e \int_{0}^{1} \frac{l n (1 + x^{2})}{1 + x} d x

Commented by mathmax by abdo last updated on 05/May/20

l e t f (a) = \int_{0}^{1} \frac{l n (a + x^{2})}{x + 1} d x w i t h a > 0

f^{^{'}} (a) = \int_{0}^{1} \frac{1}{(x + 1) (x^{2} + a)} d x l e t d e c o m p o s e F (x) = \frac{1}{(x + 1) (x^{2} + a)}

F (x) = \frac{α}{x + 1} + \frac{β x + γ}{x^{2} + a}

α = \frac{1}{a + 1}, {l i m}_{x \to + \infty} x F (x) = 0 = α + β \Rightarrow β = - \frac{1}{a + 1}

F (x) = \frac{1}{(a + 1) (x + 1)} + \frac{- \frac{1}{a + 1} x + γ}{x^{2} + a}

F (0) = \frac{1}{a} = \frac{1}{a + 1} + \frac{γ}{a} \Rightarrow 1 = \frac{a}{a + 1} + γ \Rightarrow γ = 1 - \frac{a}{a + 1} = \frac{1}{a + 1} \Rightarrow

F (x) = \frac{1}{a + 1} {\frac{1}{x + 1} + \frac{- x + 1}{x^{2} + a}} \Rightarrow

\int_{0}^{1} F (x) d x = \frac{1}{a + 1} {\int_{0}^{1} \frac{d x}{x + 1} - \frac{1}{2} \int_{0}^{1} \frac{2 x}{x^{2} + a} d x + \int_{0}^{1} \frac{d x}{x^{2} + a}}

\int_{0}^{1} \frac{d x}{x + 1} = {[l n (x + 1)]}_{0}^{1} = l n 2

\int_{0}^{1} \frac{2 x}{x^{2} + a} d x = {[l n (x^{2} + a)]}_{0}^{1} = l n (1 + a) - l n (a)

\int_{0}^{1} \frac{d x}{x^{2} + a} =_{x = \sqrt{a} t} \int_{0}^{\frac{1}{\sqrt{a}}} \frac{\sqrt{a} d t}{a (1 + t^{2})} = \frac{1}{\sqrt{a}} {[a r c t a n (t)]}_{0}^{\frac{1}{\sqrt{a}}}

= \frac{1}{\sqrt{a}} {a r c t a n (\frac{1}{\sqrt{a}})} \Rightarrow

f^{^{'}} (a) = \frac{l n (2)}{a + 1} - \frac{1}{2 (a + 1)} (l n (1 + a) - l n (a)) + \frac{1}{\sqrt{a} (a + 1)} a r c t a n (\frac{1}{\sqrt{a}}) \Rightarrow

f (a) = l n (2) \int \frac{d a}{a + 1} - \frac{1}{2} \int \frac{l n (1 + a) - l n (a)}{a + 1} d a

+ \int \frac{a r c t a n (\frac{1}{\sqrt{a}})}{\sqrt{a} (a + 1)} d a + C

\int \frac{d a}{a + 1} = l n (a + 1) + c_{0}

\int \frac{a r c t a n (\frac{1}{\sqrt{a}})}{\sqrt{a} (a + 1)} d a =_{\sqrt{a} = x} \int \frac{a r c t a n (\frac{1}{x})}{x (1 + x^{2})} (2 x) d x

= 2 \int \frac{\frac{π}{2} - a r c t a n x}{1 + x^{2}} d x = π a r c t a n (x) - 2 \int \frac{a r c t a n x}{1 + x^{2}} d x

b u t b y p a r t s \int \frac{a r c t a n x}{1 + x^{2}} d x = {a r c t a n}^{2} x - \int \frac{a r c t a n x}{1 + x^{2}} d x \Rightarrow

\int \frac{a r c t a n x}{1 + x^{2}} = \frac{1}{2} {a r c t a n}^{2} (x) + c_{1} \Rightarrow

\int \frac{a r c t a n (\frac{1}{\sqrt{a}})}{\sqrt{a} (a + 1)} d a = π a r c t a n (\sqrt{a}) - {a r c t a n}^{2} (\sqrt{a}) + c_{1}

r e s t t o f i n d \int \frac{l n (1 + a) - l n (a)}{a + 1} d a \dots . b e c o n t i n u e d \dots .