new attempt to solve qu. 37630 ∫(dx/((√x)+(√(x+1))+(√(x+2))))= [t=x+1 → dx=dt] =∫(dt/((√(t−1))+(√t)+(√(t+1))))= [((to omit the roots)),(((√a)+(√b)+(√c) must be multiplied with)),(((−(√a)−(√b)+(√c))(−(√a)+(√b)−(√c))((√a)−(√b)−(√c)))),(((1/((√a)+(√b)+(√c)))=((a^(3/2) +b^(3/2) +c^(3/2) +2(√(abc))−((a+b)(√c)+(a+c)(√b)+(b+c)(√a)))/(a^2 +b^2 +c^2 −2(ab+ac+bc))))) ] =∫((t(√(t−1))+t(√t)+t(√(t+1))+2(√(t−1))−2(√(t+1))−2(√((t−1)t(t+1))))/(3t^2 −4))dt= =∫((t(√(t−1)))/(3t^2 −4))dt+∫((t(√t))/(3t^2 −4))dt+∫((t(√(t+1)))/(3t^2 −4))dt+2∫((√(t−1))/(3t^2 −4))dt−2∫((√(t+1))/(3t^2 −4))−2∫((√((t−1)t(t+1)))/(3t^2 −4))dt I think I can solve them all except the last one so please somebody try ∫((√((t−1)t(t+1)))/(3t^2 −4))dt=? I will do the others tomorrow


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Question Number 38451 by MJS last updated on 25/Jun/18

$new attempt to solve qu . 37630$ $\int \frac{d x}{\sqrt{x} + \sqrt{x + 1} + \sqrt{x + 2}} =$ $[t = x + 1 \to d x = d t]$ $= \int \frac{d t}{\sqrt{t - 1} + \sqrt{t} + \sqrt{t + 1}} =$ $[\begin{matrix} to omit the roots \\ \sqrt{a} + \sqrt{b} + \sqrt{c} must be multiplied with \\ (- \sqrt{a} - \sqrt{b} + \sqrt{c}) (- \sqrt{a} + \sqrt{b} - \sqrt{c}) (\sqrt{a} - \sqrt{b} - \sqrt{c}) \\ \frac{1}{\sqrt{a} + \sqrt{b} + \sqrt{c}} = \frac{a^{3 / 2} + b^{3 / 2} + c^{3 / 2} + 2 \sqrt{a b c} - ((a + b) \sqrt{c} + (a + c) \sqrt{b} + (b + c) \sqrt{a})}{a^{2} + b^{2} + c^{2} - 2 (a b + a c + b c)} \end{matrix}]$ $= \int \frac{t \sqrt{t - 1} + t \sqrt{t} + t \sqrt{t + 1} + 2 \sqrt{t - 1} - 2 \sqrt{t + 1} - 2 \sqrt{(t - 1) t (t + 1)}}{3 t^{2} - 4} d t =$ $= \int \frac{t \sqrt{t - 1}}{3 t^{2} - 4} d t + \int \frac{t \sqrt{t}}{3 t^{2} - 4} d t + \int \frac{t \sqrt{t + 1}}{3 t^{2} - 4} d t + 2 \int \frac{\sqrt{t - 1}}{3 t^{2} - 4} d t - 2 \int \frac{\sqrt{t + 1}}{3 t^{2} - 4} - 2 \int \frac{\sqrt{(t - 1) t (t + 1)}}{3 t^{2} - 4} d t$ $I think I can solve them all except the last one$ $so please somebody try$ $\int \frac{\sqrt{(t - 1) t (t + 1)}}{3 t^{2} - 4} d t = ?$ $I will do the others tomorrow$
Commented by math khazana by abdo last updated on 26/Jun/18

$c h a n g e m e n t t = c h (x) g i v e$ $\int \frac{\sqrt{(t - 1) t (t + 1)}}{3 t^{2} - 4} d t = \int \frac{\sqrt{{s h}^{2} x c h x}}{3 {c h}^{2} (x) - 4} s h (x) d x$ $= \int \frac{{s h}^{2} x \sqrt{c h x}}{3 {c h}^{2} (x) - 4} d x$ $=_{\sqrt{c h x} = u} \int \frac{(u^{4} - 1) u}{3 u^{4} - 4} (c h x = u^{2} \Rightarrow x = a r g c h (u^{2})$ $\Rightarrow d x = \frac{2 u d u}{\sqrt{u^{4} - 1}} \Rightarrow$ $I = \int \frac{(u^{4} - 1)}{3 u^{4} - 4} \frac{2 u d u}{\sqrt{u^{4} - 1}}$ $= 2 \int \frac{u \sqrt{u^{4} - 1}}{3 u^{4} - 4} d u = 2 \int \frac{u \sqrt{u^{4} - 1}}{4 (u^{4} - 1) - u^{4}} d u$ $=_{\sqrt{u^{4} - 1} = x} 2 \int \frac{{(1 + x^{2})}^{\frac{1}{4}} x}{4 x^{2} - (1 + x^{2})} \frac{x}{2} {(1 + x^{2})}^{- \frac{3}{4}} (u^{4} = 1 + x^{2} \Rightarrow$ $u = {(1 + x^{2})}^{\frac{1}{4}} \Rightarrow d u = \frac{x}{2} {(1 + x^{2})}^{- \frac{3}{4}}$ $I = \int \frac{x^{2}}{\sqrt{1 + x^{2}} (3 x^{2} - 1)} d x = . . .$
	Answered by MJS last updated on 27/Jun/18

	$\int \frac{t \sqrt{t - 1}}{3 t^{2} - 4} d t =$ $[u = \sqrt{t - 1} \to d t = 2 \sqrt{t - 1} d u]$ $= 2 \int \frac{u^{4} + u^{2}}{3 u^{4} + 6 u^{2} - 1} d u = \frac{2}{3} \int (1 - \frac{3 u^{2} - 1}{3 u^{4} + 6 u^{2} - 1}) d u =$ $= \frac{2}{3} \int d u + \frac{2}{3} \int \frac{d u}{3 u^{4} + 6 u^{2} - 1} - 2 \int \frac{u^{2}}{3 u^{4} + 6 u^{2} - 1} d u =$ $\frac{2}{3} \int d u = \frac{2}{3} u = \frac{2}{3} \sqrt{t - 1}$ $\frac{2}{3} \int \frac{d u}{3 u^{4} + 6 u^{2} - 1} = \frac{2}{9} \int \frac{d u}{u^{4} + 2 u^{2} - \frac{1}{3}} =$ $= \frac{2}{9} \int \frac{d u}{(u^{2} + \frac{3 - 2 \sqrt{3}}{3}) (u^{2} + \frac{3 + 2 \sqrt{3}}{3})} =$ $= \frac{2}{9} \int \frac{A}{u^{2} + \frac{3 - 2 \sqrt{3}}{3}} d u + \frac{2}{9} \int \frac{B}{u^{2} + \frac{3 + 2 \sqrt{3}}{3}} d u =$ $= \frac{\sqrt{3}}{18} \int \frac{d u}{u^{2} + \frac{3 - 2 \sqrt{3}}{3}} - \frac{\sqrt{3}}{18} \int \frac{d u}{u^{2} + \frac{3 + 2 \sqrt{3}}{3}} =$ $[\int \frac{d v}{v^{2} - p^{2}} = \frac{1}{2 p} \ln \frac{v - p}{v + p}; \int \frac{d v}{v^{2} + q^{2}} = \frac{1}{q} \arctan \frac{v}{q}]$ $= \frac{\sqrt{3}}{36} \sqrt{3 + 2 \sqrt{3}} \ln ∣ \frac{u - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{u + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ - \frac{\sqrt{3}}{18} \sqrt{- 3 + 2 \sqrt{3}} \arctan \sqrt{- 3 + 2 \sqrt{3}} u =$ $= \frac{\sqrt{3}}{36} (\sqrt{3 + 2 \sqrt{3}} \ln ∣ \frac{\sqrt{t - 1} - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{\sqrt{t - 1} + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ - 2 \sqrt{- 3 + 2 \sqrt{3}} \arctan \sqrt{(- 3 + 2 \sqrt{3}) (t - 1)})$ $- 2 \int \frac{u^{2}}{3 u^{4} + 6 u^{2} - 1} d u = - \frac{2}{3} \int \frac{u^{2}}{u^{4} + 2 u^{2} - \frac{1}{3}} d u =$ $= - \frac{2}{3} \int \frac{u^{2}}{(u^{2} + \frac{3 - 2 \sqrt{3}}{3}) (u^{2} + \frac{3 + 2 \sqrt{3}}{3})} d u =$ $= - \frac{2}{3} \int \frac{C}{u^{2} + \frac{3 - 2 \sqrt{3}}{3}} d u - \frac{2}{3} \int \frac{D}{u^{2} + \frac{3 + 2 \sqrt{3}}{3}} d u =$ $= - \frac{2 - \sqrt{3}}{6} \int \frac{d u}{u^{2} + \frac{3 - 2 \sqrt{3}}{3}} - \frac{2 + \sqrt{3}}{6} \int \frac{d u}{u^{2} + \frac{3 + 2 \sqrt{3}}{3}} =$ $= - \frac{1}{12} (\sqrt{- 3 + 2 \sqrt{3}} \ln ∣ \frac{u - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{u + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ + 2 \sqrt{3 + 2 \sqrt{3}} \arctan \sqrt{- 3 + 2 \sqrt{3}} u) =$ $= - \frac{1}{12} (\sqrt{- 3 + 2 \sqrt{3}} \ln ∣ \frac{\sqrt{t - 1} - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{\sqrt{t - 1} + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ + 2 \sqrt{3 + 2 \sqrt{3}} \arctan \sqrt{(- 3 + 2 \sqrt{3}) (t - 1)})$ $= \frac{2}{3} \sqrt{t - 1} + \frac{\sqrt{3}}{18} \sqrt{- 3 + 2 \sqrt{3}} \ln ∣ \frac{\sqrt{t - 1} - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{\sqrt{t - 1} + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ + \frac{\sqrt{3}}{9} \sqrt{3 + 2 \sqrt{3}} \arctan \sqrt{(- 3 + 2 \sqrt{3}) (t - 1)}$
	Commented by MJS last updated on 27/Jun/18

	$\int \frac{t \sqrt{t + 1}}{3 t^{2} - 4} d t =$ $[same procedure as above]$ $= \frac{2}{3} \sqrt{t + 1} + \frac{\sqrt{3}}{18} \sqrt{3 + 2 \sqrt{3}} \ln ∣ \frac{\sqrt{t + 1} - \frac{\sqrt{3}}{3} \sqrt{3 + 2 \sqrt{3}}}{\sqrt{t + 1} + \frac{\sqrt{3}}{3} \sqrt{3 + 2 \sqrt{3}}} ∣ - \frac{\sqrt{3}}{9} \sqrt{- 3 + 2 \sqrt{3}} \arctan \sqrt{(3 + 2 \sqrt{3}) (t + 1)}$
	Commented by MJS last updated on 27/Jun/18

	$\int \frac{t \sqrt{t}}{3 t^{2} - 4} d t =$ $[ditto]$ $= \frac{2}{3} \sqrt{t} + \frac{\sqrt{6 \sqrt{3}}}{18} \ln ∣ \frac{\sqrt{t} - \frac{\sqrt{6 \sqrt{3}}}{3}}{\sqrt{t} + \frac{\sqrt{6 \sqrt{3}}}{3}} ∣ - \frac{\sqrt{6 \sqrt{3}}}{9} \arctan \frac{\sqrt{2 \sqrt{3} t}}{2}$
	Commented by MJS last updated on 27/Jun/18

	$2 \int \frac{\sqrt{t - 1}}{3 t^{2} - 4} d t =$ $= \frac{1}{6} \sqrt{- 3 + 2 \sqrt{3}} \ln ∣ \frac{\sqrt{t - 1} - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{\sqrt{t - 1} + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ - \frac{1}{3} \sqrt{3 + 2 \sqrt{3}} \arctan \sqrt{(- 3 + 2 \sqrt{3}) (t - 1)}$ $- 2 \int \frac{\sqrt{t + 1}}{3 t^{2} - 4} d t =$ $= - \frac{1}{6} \sqrt{3 + 2 \sqrt{3}} \ln ∣ \frac{\sqrt{t + 1} - \frac{\sqrt{3}}{3} \sqrt{3 + 2 \sqrt{3}}}{\sqrt{t + 1} + \frac{\sqrt{3}}{3} \sqrt{3 + 2 \sqrt{3}}} ∣ - \frac{1}{3} \sqrt{- 3 + 2 \sqrt{3}} \arctan \sqrt{(3 + 2 \sqrt{3}) (t + 1)}$
	Commented by MJS last updated on 27/Jun/18

	$\int \frac{d t}{\sqrt{t - 1} + \sqrt{t} + \sqrt{t + 1}} =$ $= \frac{2}{3} (\sqrt{t - 1} + \sqrt{t} + \sqrt{t + 1}) +$ $+ \frac{\sqrt{6 \sqrt{3}}}{18} (\ln ∣ \frac{\sqrt{t - 1} - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{\sqrt{t - 1} + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣ + \ln ∣ \frac{\sqrt{t} - \frac{\sqrt{6 \sqrt{3}}}{3}}{\sqrt{t} + \frac{\sqrt{6 \sqrt{3}}}{3}} ∣ - \ln ∣ \frac{\sqrt{t + 1} - \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}}{\sqrt{t + 1} + \frac{\sqrt{3}}{3} \sqrt{- 3 + 2 \sqrt{3}}} ∣) +$ $- \frac{\sqrt{6 \sqrt{3}}}{9} (\arctan \sqrt{(- 3 + 2 \sqrt{3}) (t - 1)} + \arctan \frac{\sqrt{2 \sqrt{3} t}}{2} + \arctan \sqrt{(3 + 2 \sqrt{3}) (t + 1)}) -$ $- 2 \int \frac{\sqrt{(t - 1) t (t + 1)}}{3 t^{2} - 4} d t + C$ $under construction$
	Commented by math khazana by abdo last updated on 27/Jun/18

	$s i r M j s y o u a r e r e a l l y t i r e d w i t h t h i s i t e g r a l$ $t h a n k y o u f o r t h i s h a r d w o r k y o u a r e r e a l l y$ $p a t i e n t . . .$
	Commented by MJS last updated on 27/Jun/18

	$this integral is my personal nemesis LOL$
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