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Question Number 118834 by bramlexs22 last updated on 20/Oct/20

$$\int \frac{x^4+1}{x^5+4x^3}dx$$

Answered by bobhans last updated on 20/Oct/20

$$Solve\int \frac{x^4+1}{x^5+4x^3}dx.$$$$solution.$$$$\psi =\int \frac{x^4+1}{x^3(x^2+4)}dx=\int \frac{x+x^{-3}}{x^2+4}dx$$$$\psi =\int \frac{x}{x^2+4}dx+\int \frac{x^{-3}}{x^2+4}dx$$$$\psi =\frac{1}{2}\ln (x^2+4)+\int \frac{dx}{x^3(x^2+4)}$$$$secondintegral{\psi }_2=\int \frac{dx}{x^3(x^2+4)}$$$$letx=2\tan \alpha$$$${\psi }_2=\int \frac{2\sec {}^2\alpha d\alpha }{8\tan {}^3\alpha \left(4\sec {}^2\alpha \right)}=\frac{1}{16}\int \frac{\cos {}^2\alpha d\left(\sin \alpha \right)}{\sin {}^3\alpha }$$$${\psi }_2=\int \left(\frac{1-\sin {}^2\alpha }{\sin {}^3\alpha }\right)d\left(\sin \alpha \right)=\int (u^{-3}-u^{-1})du$$$$=-\frac{1}{2u^2}-\ln \left(u\right)+c=-\frac{1}{2\sin {}^{2}x}-\ln \left(\sin x\right)+c$$$$Thus\psi =\frac{1}{2}\ln (x^2+4)-\ln \left(\sin x\right)-\frac{1}{2}\mathrm{cosec}{}^{2}x+c$$$$\psi =\ln \left(\frac{\sqrt{x^2+4}}{\sin x}\right)-\frac{\mathrm{cosec}{}^{2}x}{2}+c$$

Answered by MJS_new last updated on 20/Oct/20

$$\int \frac{x^4+1}{x^3(x^2+4)}dx=$$$$=\frac{17}{16}\int \frac{x}{x^2+4}dx-\frac{1}{16}\int \frac{dx}{x}+\frac{1}{4}\int \frac{dx}{x^3}=$$$$=\frac{17}{32}\ln (x^2+4)-\frac{1}{16}\ln \mid x\mid -\frac{1}{8x^2}+C$$

Commented by bobhans last updated on 20/Oct/20

$$typosir.x^4(x^2+4)=x^6+4x^4$$

Commented by MJS_new last updated on 20/Oct/20

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

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