1-dx-x-4-2x-2-9-2-arc-tan-x-3-arc-tan-x-3-arc-tan-1-5-

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} (\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 = \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