Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 131064 by shaker last updated on 01/Feb/21

Answered by Dwaipayan Shikari last updated on 01/Feb/21

(1/4)∫(1/x)((1/x^2 )−(1/(x^2 +4)))dx =−(1/(8x^2 ))−(1/(16))∫(1/x)−(x/(x^2 +4))dx =−(1/(8x^2 ))+(1/(16))log(((√(x^2 +4))/x))+C

$$\frac{\mathrm{1}}{\mathrm{4}}\int\frac{\mathrm{1}}{{x}}\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{4}}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\mathrm{1}}{{x}}−\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{16}}{log}\left(\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{{x}}\right)+{C} \\ $$

Answered by benjo_mathlover last updated on 01/Feb/21

∫ (x^(−3) /(x^2 (1+4x^(−2) ))) dx let u = 1+4x^(−2) ⇒du=−8x^(−3) dx or x^(−3) dx = −(du/8) and (4/x^2 ) = u−1 ⇒(1/x^2 ) = ((u−1)/4) then we has I = ∫ (1/u).((u−1)/4).(−(du/8)) I=−(1/(32))∫ ((u−1)/u) du =−(1/(32))(u−ln ∣u∣)+c I=−(1/(32))(((x^2 +4)/x^2 )−ln (x^2 +4)+2ln ∣x∣ )+ c

$$\:\int\:\frac{\mathrm{x}^{−\mathrm{3}} }{\mathrm{x}^{\mathrm{2}} \left(\mathrm{1}+\mathrm{4x}^{−\mathrm{2}} \right)}\:\mathrm{dx}\: \\ $$$$\:\mathrm{let}\:\mathrm{u}\:=\:\mathrm{1}+\mathrm{4x}^{−\mathrm{2}} \:\Rightarrow\mathrm{du}=−\mathrm{8x}^{−\mathrm{3}} \:\mathrm{dx}\:\mathrm{or}\:\mathrm{x}^{−\mathrm{3}} \:\mathrm{dx}\:=\:−\frac{\mathrm{du}}{\mathrm{8}} \\ $$$$\mathrm{and}\:\frac{\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{u}−\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{u}−\mathrm{1}}{\mathrm{4}} \\ $$$$\mathrm{then}\:\mathrm{we}\:\mathrm{has}\:\mathrm{I}\:=\:\int\:\frac{\mathrm{1}}{\mathrm{u}}.\frac{\mathrm{u}−\mathrm{1}}{\mathrm{4}}.\left(−\frac{\mathrm{du}}{\mathrm{8}}\right) \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{32}}\int\:\frac{\mathrm{u}−\mathrm{1}}{\mathrm{u}}\:\mathrm{du}\:=−\frac{\mathrm{1}}{\mathrm{32}}\left(\mathrm{u}−\mathrm{ln}\:\mid\mathrm{u}\mid\right)+\mathrm{c} \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\mathrm{32}}\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{2}} }−\mathrm{ln}\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)+\mathrm{2ln}\:\mid\mathrm{x}\mid\:\right)+\:\mathrm{c} \\ $$

Answered by Ar Brandon last updated on 01/Feb/21

I=∫(dx/(x^3 (x^2 +4))) , x=2tanθ =∫((2sec^2 θ)/(8tan^3 θ(4tan^2 θ+4)))dθ =(1/(16))∫(dθ/(tan^3 θ))=(1/(16))∫(((1−sin^2 θ))/(sin^3 θ))d(sinθ) =−(1/(32sin^2 θ))−((ln∣sinθ∣)/(16))+C , θ=tan^(−1) ((x/2))

$$\mathcal{I}=\int\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}\right)}\:,\:\mathrm{x}=\mathrm{2tan}\theta \\ $$$$\:\:\:=\int\frac{\mathrm{2sec}^{\mathrm{2}} \theta}{\mathrm{8tan}^{\mathrm{3}} \theta\left(\mathrm{4tan}^{\mathrm{2}} \theta+\mathrm{4}\right)}\mathrm{d}\theta \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\mathrm{d}\theta}{\mathrm{tan}^{\mathrm{3}} \theta}=\frac{\mathrm{1}}{\mathrm{16}}\int\frac{\left(\mathrm{1}−\mathrm{sin}^{\mathrm{2}} \theta\right)}{\mathrm{sin}^{\mathrm{3}} \theta}\mathrm{d}\left(\mathrm{sin}\theta\right) \\ $$$$\:\:\:=−\frac{\mathrm{1}}{\mathrm{32sin}^{\mathrm{2}} \theta}−\frac{\mathrm{ln}\mid\mathrm{sin}\theta\mid}{\mathrm{16}}+\mathcal{C}\:,\:\theta=\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{2}}\right) \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com