Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

$$\mathrm{yes},\:\mathrm{thank}\:\mathrm{you} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com