find A_n = ∫_(1/n) ^1 x(√x)arctan(x+(1/x))dx then calculate lim_(n→+∞) A


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 37813 by prof Abdo imad last updated on 17/Jun/18

$f i n d A_{n} = \int_{\frac{1}{n}}^{1} x \sqrt{x} a r c t a n (x + \frac{1}{x}) d x$ $t h e n c a l c u l a t e {l i m}_{n \to + \infty} A_{n} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Jun/18

	$\int {t a n}^{- 1} (x + \frac{1}{x}) \times x^{\frac{3}{2}} d x$ ${t a n}^{- 1} (x + \frac{1}{x}) \times \frac{x^{\frac{5}{2}}}{\frac{5}{2}} - \int \frac{1}{1 + x^{2} + 2 + \frac{1}{x^{2}}} \times \frac{x^{\frac{5}{2}}}{\frac{5}{2}} d x$ $d o - \frac{2}{5} \int \frac{x^{\frac{5}{2}}}{x^{2} + \frac{1}{x^{2}} + 3} d x$ $d 0 - \frac{2}{5} \int \frac{x^{\frac{9}{2}}}{x^{4} + 3 x^{2} + 1} d x$ $x = t^{2} d x = 2 t d t$ $d o - \frac{2}{5} \int \frac{t^{2 \times \frac{9}{2}} \times 2 t d t}{t^{8} + 3 t^{4} + 1}$ $d o - \frac{4}{5} \int \frac{t^{10}}{t^{8} + 3 t^{4} + 1} d t$ $d o - \frac{4}{5} \int \frac{t^{10} + 3 t^{6} + t^{2} - 3 t^{6} - t^{2}}{t^{8} + 3 t^{4} + 1} d t$ $d o - \frac{4}{5} \int t^{2} d t + \frac{4}{5} \int \frac{3 t^{6} + t^{2}}{t^{8} + 3 t^{4} + 1} d t$ $d o - \frac{4}{5} \int t^{2} d t + \frac{4}{5} \int \frac{3 t^{2} + \frac{1}{t^{2}}}{t^{4} + 3 + \frac{1}{t^{4}}}$ $d o - \frac{4}{5} \int t^{2} d t + \frac{4}{5} \int \frac{3 (t^{2} + \frac{1}{t^{2}}) - \frac{2}{t^{2}}}{{(t^{2} + \frac{1}{t^{2}})}^{2} + 1}$ $c o n t d$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum