find-pi-4-4-pi-1-1-x-2-arctanx-dx-

Question Number 33328 by prof Abdo imad last updated on 14/Apr/18

f i n d \int_{\frac{π}{4}}^{\frac{4}{π}} (1 + \frac{1}{x^{2}}) a r c t a n x d x

Commented by math khazana by abdo last updated on 19/Apr/18

l e t p u t I = \int_{\frac{π}{4}}^{\frac{4}{π}} (1 + \frac{1}{x^{2}}) a r c t a n x d x . l e t i n t e g r a t e b y

p a r t s u^{^{'}} = 1 + \frac{1}{x^{2}} a n d v = a r c t a n x

I = {[(1 - \frac{1}{x}) a r c t a n x]}_{\frac{π}{4}}^{\frac{4}{π}} - \int_{\frac{π}{4}}^{\frac{4}{π}} (1 - \frac{1}{x}) \frac{d x}{1 + x^{2}}

= (1 - \frac{π}{4}) a r c t a n (\frac{4}{π}) - (1 - \frac{4}{π}) - \int_{\frac{π}{4}}^{\frac{4}{π}} \frac{d x}{1 + x^{2}}

+ \int_{\frac{π}{4}}^{\frac{4}{π}} \frac{d x}{x (1 + x^{2})} b u t

\int_{\frac{π}{4}}^{\frac{4}{π}} \frac{d x}{1 + x^{2}} = a r c t a n (\frac{4}{π}) - a r c t a n (\frac{π}{4})

= \frac{π}{2} - 1 - 1 = \frac{π}{2} - 2 l e t ? d e c o m p o s e

F (x) = \frac{1}{x (1 + x^{2})} = \frac{a}{x} + \frac{b x + c}{1 + x^{2}}

a = {l i m}_{x \to 0} x F (x) = 1

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

F (x) = \frac{1}{x} + \frac{- x + c}{1 + x^{2}} w e l o o k t b a t c = 0 \Rightarrow

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

\int_{\frac{π}{4}}^{\frac{4}{π}} \frac{d x}{x (1 + x^{2})} = \int_{\frac{π}{4}}^{\frac{4}{π}} (\frac{1}{x} - \frac{x}{1 + x^{2}}) d x

= {[l n (x) - \frac{1}{2} l n (1 + x^{2})]}_{\frac{π}{4}}^{\frac{4}{π}} = {[l n (\frac{x}{\sqrt{1 + x^{2}}})]}_{\frac{π}{4}}^{\frac{4}{π}}

= l n (\frac{4}{π \sqrt{1 + \frac{16}{π^{2}}}}) - l n (\frac{π}{4 \sqrt{1 + \frac{π^{2}}{16}}})

I = (1 - \frac{π}{4}) (\frac{π}{2} - 1) + 1 + \frac{4}{π} - \frac{π}{2}

+ l n (\frac{4}{π \sqrt{1 + \frac{16}{π^{2}}}}) - l n (\frac{π}{4 \sqrt{1 + \frac{π^{2}}{16}}}) .