find-1-arctan-x-x-2-

Question Number 27805 by abdo imad last updated on 15/Jan/18

f i n d \int_{1}^{\propto} \frac{a r c t a n (α x)}{x^{2}} .

Commented by abdo imad last updated on 16/Jan/18

l e t p u t I = \int_{1}^{\propto} \frac{a r c t a n (α x)}{x^{2}} d x i n t e g r a t e p e r p a r t s

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

I = a r t a n (α) + α \int_{1}^{+ \propto} \frac{d x}{x (1 + α^{2} x^{2})} w e u s e t h e c h . x = \frac{1}{α} t

\int_{1}^{+ \propto} \frac{d x}{x (1 + α^{2} x^{2})} = \int_{α}^{+ \propto} \frac{\frac{1}{α} d t}{\frac{1}{α} t (1 + t^{2})} = \int_{α}^{+ \propto} \frac{d t}{t (1 + t^{2})}

b u t \frac{1}{t (1 + t^{2})} = \frac{1}{t} - \frac{t}{1 + t^{2}} \Rightarrow \int \frac{d t}{t (1 + t^{2})} = l n / t / - \frac{1}{2} l n (1 + t^{2}) + k

= l n / \frac{t}{\sqrt{1 + t^{2}}} / s o \int_{α}^{+ \propto} \frac{d t}{t (1 + t^{2})} = {[l n / \frac{t}{\sqrt{1 + t^{2}}} /]}_{α}^{+ \propto}

= - l n / \frac{α}{\sqrt{1 + α^{2}}} /

I = a r t a n (α) - α l n / \frac{α}{\sqrt{1 + α^{2}}} / .