calculate-0-1-arctan-x-2-x-dx-

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}}})

\dots

\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 .