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Question Number 108891 by bobhans last updated on 20/Aug/20

   ((bobHans)/∦)  ∫ (((x^2 −2) dx)/((x^4 +5x^2 +4) arc tan (((x^2 +2)/x))))

$$\:\:\:\frac{\boldsymbol{{bob}}\mathbb{H}{ans}}{\nparallel} \\ $$$$\int\:\frac{\left({x}^{\mathrm{2}} −\mathrm{2}\right)\:{dx}}{\left({x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}\right)\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}}\right)} \\ $$

Answered by bemath last updated on 20/Aug/20

   ((∧BeMath∧)/(⋰⋱⋰⋱⋰⋱))  let w = arc tan (((x^2 +2)/x))  (dw/dx) = ((d(((x^2 +2)/x)))/(1+(((x^2 +2)/x))^2 )) = (x^2 /(x^4 +5x^2 +4)) .(1−(2/x^2 ))  = (x^2 /(x^4 +5x^2 +4)) . ((x^2 −2)/x^2 ) = ((x^2 −2)/(x^4 +5x^2 +4))  I = ∫ (dw/w) = ln ∣w∣ + c = ln ∣ arc tan (((x^2 +2)/x))∣ + c

$$\:\:\:\frac{\wedge\mathcal{B}{e}\mathcal{M}{ath}\wedge}{\iddots\ddots\iddots\ddots\iddots\ddots} \\ $$$${let}\:{w}\:=\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}}\right) \\ $$$$\frac{{dw}}{{dx}}\:=\:\frac{{d}\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}}\right)}{\mathrm{1}+\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}}\right)^{\mathrm{2}} }\:=\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}}\:.\left(\mathrm{1}−\frac{\mathrm{2}}{{x}^{\mathrm{2}} }\right) \\ $$$$=\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}}\:.\:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{2}} }\:=\:\frac{{x}^{\mathrm{2}} −\mathrm{2}}{{x}^{\mathrm{4}} +\mathrm{5}{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${I}\:=\:\int\:\frac{{dw}}{{w}}\:=\:\mathrm{ln}\:\mid{w}\mid\:+\:{c}\:=\:\mathrm{ln}\:\mid\:\mathrm{arc}\:\mathrm{tan}\:\left(\frac{{x}^{\mathrm{2}} +\mathrm{2}}{{x}}\right)\mid\:+\:{c} \\ $$

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