(1) ∫ (dx/(x^4 +2x^2 +9)) ? (2) arc tan ((x/3))−arc tan ((x/3))= arc tan ((1/5))


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Question Number 121500 by bramlexs22 last updated on 08/Nov/20

$(1) \int \frac{dx}{x^{4} + {2 x}^{2} + 9} ?$ $(2) arc \tan (\frac{x}{3}) - arc \tan (\frac{x}{3}) = arc \tan (\frac{1}{5})$
Commented by benjo_mathlover last updated on 08/Nov/20
$i think it should be (2) tan^(−1) ((x/3))−tan^(−1) ((x/2))= tan^(−1) ((1/5)) let x = 6ψ ⇒ tan^(−1) (2ψ)−tan^(−1) (3ψ)=tan^(−1) ((1/5)) ⇔ ((2ψ−3ψ)/(1+6ψ^2 )) = (1/5) ; 6ψ^2 +5ψ+1=0 (2ψ+1)(3ψ+1)=0 → { (((1/6)x=−(1/2);x =−3)),(((1/6)x=−(1/3); x=−2)) :}$
$i think it should be$ $(2) \tan^{- 1} (\frac{x}{3}) - \tan^{- 1} (\frac{x}{2}) = \tan^{- 1} (\frac{1}{5})$ $let x = 6 ψ \Rightarrow \tan^{- 1} (2 ψ) - \tan^{- 1} (3 ψ) = \tan^{- 1} (\frac{1}{5})$ $\Leftrightarrow \frac{2 ψ - 3 ψ}{1 + 6 ψ^{2}} = \frac{1}{5}; 6 ψ^{2} + 5 ψ + 1 = 0$ $(2 ψ + 1) (3 ψ + 1) = 0 \to {\begin{cases} \frac{1}{6} x = - \frac{1}{2}; x = - 3 \\ \frac{1}{6} x = - \frac{1}{3}; x = - 2 \end{cases}$
	Answered by Ar Brandon last updated on 09/Nov/20
	$1\I=∫(dx/(x^4 +2x^2 +9))=(1/6)∫(((x^2 +3)−(x^2 −3))/(x^4 +2x^2 +9))dx 6I=∫((x^2 +3)/(x^4 +2x^2 +9))dx−∫((x^2 −3)/(x^4 +2x^2 +9))dx =∫((1+(3/x^2 ))/(x^2 +2+(9/x^2 )))dx−∫((1−(3/x^2 ))/(x^2 +2+(9/x^2 )))dx =∫((1+(3/x^2 ))/((x−(3/x))^2 +8))dx−∫((1−(3/x^2 ))/((x+(3/x))^2 −4))dx =∫(du/(u^2 +8))−∫(dv/(v^2 −4))=(1/(2(√2)))Arctan((u/(2(√2))))+(1/2)Arctanh((v/2))+C$
	$1 ∖ I = \int \frac{dx}{x^{4} + {2 x}^{2} + 9} = \frac{1}{6} \int \frac{(x^{2} + 3) - (x^{2} - 3)}{x^{4} + {2 x}^{2} + 9} dx$ $6 I = \int \frac{x^{2} + 3}{x^{4} + {2 x}^{2} + 9} dx - \int \frac{x^{2} - 3}{x^{4} + {2 x}^{2} + 9} dx$ $= \int \frac{1 + \frac{3}{x^{2}}}{x^{2} + 2 + \frac{9}{x^{2}}} dx - \int \frac{1 - \frac{3}{x^{2}}}{x^{2} + 2 + \frac{9}{x^{2}}} dx$ $= \int \frac{1 + \frac{3}{x^{2}}}{{(x - \frac{3}{x})}^{2} + 8} dx - \int \frac{1 - \frac{3}{x^{2}}}{{(x + \frac{3}{x})}^{2} - 4} dx$ $= \int \frac{du}{u^{2} + 8} - \int \frac{dv}{v^{2} - 4} = \frac{1}{2 \sqrt{2}} Arctan (\frac{u}{2 \sqrt{2}}) + \frac{1}{2} Arctanh (\frac{v}{2}) + C$
	Answered by MJS_new last updated on 09/Nov/20

	$\int \frac{d x}{x^{4} + 2 x^{2} + 9} = \int \frac{d x}{(x^{2} - 2 x + 3) (x^{2} + 2 x + 3)} =$ $= \frac{1}{12} \int (\frac{x + 2}{x^{2} + 2 x + 3} - \frac{x - 2}{x^{2} - 2 x + 3}) d x =$ $= \frac{1}{12} \int \frac{d x}{x^{2} + 2 x + 3} + \frac{1}{24} \int \frac{2 x + 2}{x^{2} + 2 x + 3} d x +$ $+ \frac{1}{12} \int \frac{d x}{x^{2} - 2 x + 3} - \frac{1}{24} \int \frac{2 x - 2}{x^{2} - 2 x + 3} d x =$ $= \frac{\sqrt{2}}{24} \arctan \frac{\sqrt{2} (x + 1)}{2} + \frac{1}{24} \ln (x^{2} + 2 x + 3) +$ $+ \frac{\sqrt{2}}{24} \arctan \frac{\sqrt{2} (x - 1)}{2} - \frac{1}{24} \ln (x^{2} - 2 x + 3) + C$
	Answered by TANMAY PANACEA last updated on 09/Nov/20

	$1) \int \frac{\frac{1}{x^{2}}}{x^{2} + \frac{9}{x^{2}} + 2} d x$ $\frac{1}{6} \int \frac{(1 + \frac{3}{x^{2}}) - (1 - \frac{3}{x^{2}})}{x^{2} + \frac{9}{x^{2}} + 2} d x$ $\frac{1}{6} \int \frac{d (x - \frac{3}{x})}{{(x - \frac{3}{x})}^{2} + 8} - \frac{1}{6} \int \frac{d (x + \frac{3}{x})}{{(x + \frac{3}{x})}^{2} - 4}$ $\frac{1}{6} \times \frac{1}{\sqrt{8}} {t a n}^{- 1} (\frac{x - \frac{3}{x}}{\sqrt{8}}) - \frac{1}{6} \times \frac{1}{2 \times 2} l n (\frac{x - \frac{3}{x} - 2}{x - \frac{3}{x} + 2}) + c$
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