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x-4-1-x-5-4x-3-dx-

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Question Number 118834 by bramlexs22 last updated on 20/Oct/20
∫ ((x^4 +1)/(x^5 +4x^3 )) dx
$$\:\:\int\:\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{3}} }\:{dx}\: \\ $$
Answered by bobhans last updated on 20/Oct/20
Solve ∫ ((x^4 +1)/(x^5 +4x^3 )) dx. solution. ψ = ∫ ((x^4 +1)/(x^3 (x^2 +4))) dx = ∫ ((x+x^(−3) )/(x^2 +4)) dx ψ=∫ (x/(x^2 +4)) dx + ∫ (x^(−3) /(x^2 +4)) dx ψ= (1/2)ln (x^2 +4)+∫ (dx/(x^3 (x^2 +4))) second integral ψ_2 = ∫ (dx/(x^3 (x^2 +4))) let x = 2tan α ψ_2 = ∫ ((2sec^2 α dα)/(8tan^3 α(4sec^2 α))) = (1/(16))∫ ((cos^2 α d(sin α))/(sin^3 α)) ψ_2 = ∫ (((1−sin^2 α)/(sin^3 α))) d(sin α)= ∫ (u^(−3) −u^(−1) )du =−(1/(2u^2 ))−ln (u)+ c = −(1/(2sin^2 x))−ln (sin x) + c Thus ψ = (1/2)ln (x^2 +4)−ln (sin x)−(1/2)cosec^2 x + c ψ= ln (((√(x^2 +4))/(sin x)))−((cosec^2 x)/2) + c
$$\:{Solve}\:\int\:\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{5}} +\mathrm{4}{x}^{\mathrm{3}} }\:{dx}. \\ $$$${solution}. \\ $$$$\:\psi\:=\:\int\:\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)}\:{dx}\:=\:\int\:\frac{{x}+{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{4}}\:{dx} \\ $$$$\psi=\int\:\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}}\:{dx}\:+\:\int\:\frac{{x}^{−\mathrm{3}} }{{x}^{\mathrm{2}} +\mathrm{4}}\:{dx} \\ $$$$\psi=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}\right)+\int\:\frac{{dx}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${second}\:{integral}\:\psi_{\mathrm{2}} =\:\int\:\frac{{dx}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)} \\ $$$${let}\:{x}\:=\:\mathrm{2tan}\:\alpha\: \\ $$$$\psi_{\mathrm{2}} \:=\:\int\:\frac{\mathrm{2sec}\:^{\mathrm{2}} \alpha\:{d}\alpha}{\mathrm{8tan}\:^{\mathrm{3}} \alpha\left(\mathrm{4sec}\:^{\mathrm{2}} \alpha\right)}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\:\frac{\mathrm{cos}\:^{\mathrm{2}} \alpha\:{d}\left(\mathrm{sin}\:\alpha\right)}{\mathrm{sin}\:^{\mathrm{3}} \alpha} \\ $$$$\psi_{\mathrm{2}} =\:\int\:\left(\frac{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \alpha}{\mathrm{sin}\:^{\mathrm{3}} \alpha}\right)\:{d}\left(\mathrm{sin}\:\alpha\right)=\:\int\:\left({u}^{−\mathrm{3}} −{u}^{−\mathrm{1}} \right){du} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}{u}^{\mathrm{2}} }−\mathrm{ln}\:\left({u}\right)+\:{c}\:=\:−\frac{\mathrm{1}}{\mathrm{2sin}\:^{\mathrm{2}} {x}}−\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)\:+\:{c} \\ $$$${Thus}\:\psi\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}\right)−\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cosec}\:^{\mathrm{2}} {x}\:+\:{c} \\ $$$$\psi=\:\mathrm{ln}\:\left(\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{sin}\:{x}}\right)−\frac{\mathrm{cosec}\:^{\mathrm{2}} {x}}{\mathrm{2}}\:+\:{c}\: \\ $$
Answered by MJS_new last updated on 20/Oct/20
∫((x^4 +1)/(x^3 (x^2 +4)))dx= =((17)/(16))∫(x/(x^2 +4))dx−(1/(16))∫(dx/x)+(1/4)∫(dx/x^3 )= =((17)/(32))ln (x^2 +4) −(1/(16))ln ∣x∣ −(1/(8x^2 ))+C
$$\int\frac{{x}^{\mathrm{4}} +\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)}{dx}= \\ $$$$=\frac{\mathrm{17}}{\mathrm{16}}\int\frac{{x}}{{x}^{\mathrm{2}} +\mathrm{4}}{dx}−\frac{\mathrm{1}}{\mathrm{16}}\int\frac{{dx}}{{x}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dx}}{{x}^{\mathrm{3}} }= \\ $$$$=\frac{\mathrm{17}}{\mathrm{32}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{4}\right)\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\mid{x}\mid\:−\frac{\mathrm{1}}{\mathrm{8}{x}^{\mathrm{2}} }+{C} \\ $$
Commented by bobhans last updated on 20/Oct/20
typo sir. x^4 (x^2 +4)=x^6 +4x^4
$${typo}\:{sir}.\:{x}^{\mathrm{4}} \left({x}^{\mathrm{2}} +\mathrm{4}\right)={x}^{\mathrm{6}} +\mathrm{4}{x}^{\mathrm{4}} \\ $$
Commented by MJS_new last updated on 20/Oct/20
yes, thank you
$$\mathrm{yes},\:\mathrm{thank}\:\mathrm{you} \\ $$

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