calculate ∫_0 ^1 arctan(x^2 −x)dx


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 55994 by maxmathsup by imad last updated on 07/Mar/19

$c a l c u l a t e \int_{0}^{1} a r c t a n (x^{2} - x) d x$
	Answered by MJS last updated on 08/Mar/19

	$\int \arctan (x^{2} - x) d x =$ $[t = x^{2} - x \to d x = \frac{d t}{2 x - 1}]$ $= \int \frac{\arctan t}{\sqrt{4 t + 1}} d t =$ $= \frac{\sqrt{4 t + 1}}{2} \arctan t - \frac{1}{2} \int \frac{\sqrt{4 t + 1}}{t^{2} + 1} d t$ $- \frac{1}{2} \int \frac{\sqrt{4 t + 1}}{t^{2} + 1} d t =$ $[u = \sqrt{4 t + 1} \to d t = \frac{\sqrt{4 t + 1}}{2} d u]$ $= - 4 \int \frac{u^{2}}{u^{4} - 2 u^{2} + 17} d u =$ $= - 4 \int \frac{u^{2}}{(u^{2} - \sqrt{2 + 2 \sqrt{17}} u + \sqrt{17}) (u^{2} + \sqrt{2 + 2 \sqrt{17}} u + \sqrt{17})} d u =$ $[a = \sqrt{2 + 2 \sqrt{17}}; b = \sqrt{17}]$ $= - 4 \int \frac{u^{2}}{(u^{2} - a u + b) (u^{2} + a u + b)} d u =$ $= \frac{2}{a} (\int \frac{u}{u^{2} + a u + b} d u - \int \frac{u}{u^{2} - a u + b} d u) =$ $= \frac{1}{a} \ln \frac{u^{2} + a u + b}{u^{2} - a u + b} - \frac{2}{\sqrt{4 b - a^{2}}} (\arctan \frac{2 u - a}{\sqrt{4 b - a^{2}}} + \arctan \frac{2 u + a}{\sqrt{4 b - a^{2}}})$ $. . .$ $\underset{0}{\int^{1}} \arctan (x^{2} - x) d x =$ $= \frac{\sqrt{- 2 + 2 \sqrt{17}}}{4} \ln \frac{5 + \sqrt{17} + \sqrt{26 + 10 \sqrt{17}}}{8} - \frac{\sqrt{2 + 2 \sqrt{17}}}{2} \arctan \frac{\sqrt{2 + 2 \sqrt{17}}}{4} \approx$ $\approx - . 164355$
	Commented by turbo msup by abdo last updated on 08/Mar/19

	$t h a n k s s i r .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum