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Evaluate-3x-2-5-x-4-6x-2-25-dx-

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Question Number 117446 by Ar Brandon last updated on 11/Oct/20
Evaluate ∫((3x^2 −5)/(x^4 +6x^2 +25))dx
$$\mathrm{Evaluate}\:\int\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{d}{x} \\ $$
Answered by Ar Brandon last updated on 11/Oct/20
I=∫((3x^2 −5)/(x^4 +6x^2 +25))dx =∫(((x^2 −5)+(x^2 +5)+(x^2 −5))/(x^4 +6x^2 +25))dx=∫(((x^2 +5)+2(x^2 −5))/(x^4 +6x^2 +25))dx =∫((x^2 +5)/(x^4 +6x^2 +25))dx+2∫((x^2 −5)/(x^4 +6x^2 +25))dx =∫((1+(5/x^2 ))/(x^2 +6+((25)/x^2 )))dx+2∫((1−(5/x^2 ))/(x^2 +6+((25)/x^2 )))dx =∫((1+(5/x^2 ))/((x−(5/x))^2 +16))dx+2∫((1−(5/x^2 ))/((x+(5/x))^2 −4))dx =∫(du/(u^2 +16))+2∫(dv/(v^2 −4))=(1/4)Arctan((u/4))−(2/2)Arctanh((v/2))+C =(1/4)Arctan[((x^2 −5)/(4x))]−Arctanh[((x^2 +5)/(2x))]+C
$$\mathcal{I}=\int\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{5}}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{d}{x} \\ $$$$\:\:\:=\int\frac{\left({x}^{\mathrm{2}} −\mathrm{5}\right)+\left({x}^{\mathrm{2}} +\mathrm{5}\right)+\left({x}^{\mathrm{2}} −\mathrm{5}\right)}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{d}{x}=\int\frac{\left({x}^{\mathrm{2}} +\mathrm{5}\right)+\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{5}\right)}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{d}{x} \\ $$$$\:\:\:=\int\frac{{x}^{\mathrm{2}} +\mathrm{5}}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{d}{x}+\mathrm{2}\int\frac{{x}^{\mathrm{2}} −\mathrm{5}}{{x}^{\mathrm{4}} +\mathrm{6}{x}^{\mathrm{2}} +\mathrm{25}}\mathrm{d}{x} \\ $$$$\:\:\:=\int\frac{\mathrm{1}+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\mathrm{6}+\frac{\mathrm{25}}{{x}^{\mathrm{2}} }}\mathrm{d}{x}+\mathrm{2}\int\frac{\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\mathrm{6}+\frac{\mathrm{25}}{{x}^{\mathrm{2}} }}\mathrm{d}{x} \\ $$$$\:\:\:=\int\frac{\mathrm{1}+\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}{\left({x}−\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} +\mathrm{16}}\mathrm{d}{x}+\mathrm{2}\int\frac{\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}{\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}\mathrm{d}{x} \\ $$$$\:\:\:=\int\frac{\mathrm{du}}{\mathrm{u}^{\mathrm{2}} +\mathrm{16}}+\mathrm{2}\int\frac{\mathrm{dv}}{\mathrm{v}^{\mathrm{2}} −\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{Arctan}\left(\frac{\mathrm{u}}{\mathrm{4}}\right)−\frac{\mathrm{2}}{\mathrm{2}}\mathrm{Arctanh}\left(\frac{\mathrm{v}}{\mathrm{2}}\right)+\mathcal{C} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{4}}\mathrm{Arctan}\left[\frac{{x}^{\mathrm{2}} −\mathrm{5}}{\mathrm{4}{x}}\right]−\mathrm{Arctanh}\left[\frac{{x}^{\mathrm{2}} +\mathrm{5}}{\mathrm{2}{x}}\right]+\mathcal{C} \\ $$
Answered by TANMAY PANACEA last updated on 11/Oct/20
∫((3−(5/x^2 ))/(x^2 +((25)/x^2 )+6))dx ∫((1−(5/x^2 )+2)/(x^2 +((25)/x^2 )+6))dx ∫((d(x+(5/x)))/((x+(5/x))^2 −10+6))+∫((2x^2 )/(x^4 +25+6x^2 )) ∫((d(x+(5/x)))/((x+(5/x))^2 −4))+∫(((d/dx)(x+(5/x))+(d/dx)(x−(5/x)))/(x^2 +((25)/x^2 )+6)) now use formula ∫((d(x+(5/x)))/((x+(5/x))^2 −4))+∫((d(x+(5/x)))/((x+(5/x))^2 −10+6))+∫((d(x−(5/x)))/((x−(5/x))^2 +2+6)) 2∫(dt/(t^2 −4))+∫(dk/(k^2 +8))
$$\int\frac{\mathrm{3}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }}{{x}^{\mathrm{2}} +\frac{\mathrm{25}}{{x}^{\mathrm{2}} }+\mathrm{6}}{dx} \\ $$$$\int\frac{\mathrm{1}−\frac{\mathrm{5}}{{x}^{\mathrm{2}} }+\mathrm{2}}{{x}^{\mathrm{2}} +\frac{\mathrm{25}}{{x}^{\mathrm{2}} }+\mathrm{6}}{dx} \\ $$$$\int\frac{{d}\left({x}+\frac{\mathrm{5}}{{x}}\right)}{\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} −\mathrm{10}+\mathrm{6}}+\int\frac{\mathrm{2}{x}^{\mathrm{2}} }{{x}^{\mathrm{4}} +\mathrm{25}+\mathrm{6}{x}^{\mathrm{2}} } \\ $$$$\int\frac{{d}\left({x}+\frac{\mathrm{5}}{{x}}\right)}{\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}+\int\frac{\frac{{d}}{{dx}}\left({x}+\frac{\mathrm{5}}{{x}}\right)+\frac{{d}}{{dx}}\left({x}−\frac{\mathrm{5}}{{x}}\right)}{{x}^{\mathrm{2}} +\frac{\mathrm{25}}{{x}^{\mathrm{2}} }+\mathrm{6}} \\ $$$${now}\:{use}\:{formula} \\ $$$$\int\frac{{d}\left({x}+\frac{\mathrm{5}}{{x}}\right)}{\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} −\mathrm{4}}+\int\frac{{d}\left({x}+\frac{\mathrm{5}}{{x}}\right)}{\left({x}+\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} −\mathrm{10}+\mathrm{6}}+\int\frac{{d}\left({x}−\frac{\mathrm{5}}{{x}}\right)}{\left({x}−\frac{\mathrm{5}}{{x}}\right)^{\mathrm{2}} +\mathrm{2}+\mathrm{6}} \\ $$$$\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{2}} −\mathrm{4}}+\int\frac{{dk}}{{k}^{\mathrm{2}} +\mathrm{8}} \\ $$$$ \\ $$

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