calculate-1-arctan-3-x-x-2-dx-

Question Number 84570 by msup trace by abdo last updated on 14/Mar/20

c a l c u l a t e \int_{1}^{+ \infty} \frac{a r c t a n (\frac{3}{x})}{x^{2}} d x

Commented by mathmax by abdo last updated on 15/Mar/20

l e t I = \int_{1}^{+ \infty} \frac{a r c t a n (\frac{3}{x})}{x^{2}} d x c h a n g e m e n t \frac{3}{x} = t g i v e \frac{x}{3} = \frac{1}{t} \Rightarrow

x = \frac{3}{t} \Rightarrow I = \int_{3}^{0} \frac{a r c t a n (t)}{{(\frac{3}{t})}^{2}} (- \frac{3}{t^{2}}) d t

= \frac{1}{3} \int_{0}^{3} a r c t a n t d t =_{b y p a r t s} = \frac{1}{3} {{[t a r c t a n t]}_{0}^{3} - \int_{0}^{3} t \frac{d t}{1 + t^{2}}}

3 I = 3 a r c t a n (3) - \frac{1}{2} {[l n (1 + t^{2})]}_{0}^{3}

= 3 a r c t a n (3) - \frac{1}{2} l n (10) \Rightarrow I = a r c t a n (3) - \frac{1}{6} l n (10)

Answered by M±th+et£s last updated on 14/Mar/20

y = \frac{3}{x} d y = \frac{- 3}{x^{2}}

I = \int_{3}^{0} {t a n}^{- 1} (y) \frac{d y}{- 3} = \frac{1}{3} \int_{0}^{3} {t a n}^{- 1} y d y

= \frac{1}{3} {[y {t a n}^{- 1} y - \frac{1}{2} l n (y^{2} + 1)]}_{0}^{3}

\frac{1}{3} ({t a n}^{- 1} (x) - \frac{1}{2} l n (10))