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Question Number 132333 by liberty last updated on 13/Feb/21

$$\mathrm{very}\mathrm{nice}\mathrm{integral}$$$$\int \frac{{4\mathrm{x}}^3+{4\mathrm{x}}^2+4\mathrm{x}+3}{({\mathrm{x}}^2+1){({\mathrm{x}}^2+\mathrm{x}+1)}^2}\mathrm{dx}?$$$$$$

Answered by EDWIN88 last updated on 13/Feb/21

$$\mathrm{well}...\mathrm{let}\mathrm{me}\mathrm{try}$$$$\mathrm{Ostrogradsky}\mathrm{method}$$$$\mathrm{the}\mathrm{integral}\mathrm{has}\mathrm{form}$$$$\int \frac{{4\mathrm{x}}^3+{4\mathrm{x}}^2+4\mathrm{x}+3}{({\mathrm{x}}^2+1){({\mathrm{x}}^2+\mathrm{x}+1)}^2}\mathrm{dx}=\frac{\mathrm{ax}+\mathrm{b}}{{\mathrm{x}}^2+\mathrm{x}+1}+\int \frac{{\mathrm{cx}}^3+{\mathrm{dx}}^2+\mathrm{ex}+\mathrm{f}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+\mathrm{x}+1)}\mathrm{dx}$$$$\mathrm{differentiating}\mathrm{both}\mathrm{sides}$$$$\frac{{4\mathrm{x}}^3+{4\mathrm{x}}^2+4\mathrm{x}+3}{({\mathrm{x}}^2+1){({\mathrm{x}}^2+\mathrm{x}+1)}^2}=\frac{\mathrm{a}({\mathrm{x}}^2+\mathrm{x}+1)-(\mathrm{ax}+\mathrm{b})(2\mathrm{x}+1)}{{({\mathrm{x}}^2+\mathrm{x}+1)}^2}+\frac{{\mathrm{cx}}^3+{\mathrm{dx}}^2+\mathrm{ex}+\mathrm{f}}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+\mathrm{x}+1)}$$$$\mathrm{after}\mathrm{solving}\mathrm{the}\mathrm{coefficients}\mathrm{we}\mathrm{get}$$$$\mathrm{a}=1,\mathrm{b}=-1,\mathrm{c}=0,\mathrm{d}=1,\mathrm{e}=1,\mathrm{f}=1$$$$\mathrm{then}\mathrm{I}=\frac{\mathrm{x}-1}{{\mathrm{x}}^2+\mathrm{x}+1}+\int \frac{{\mathrm{x}}^2+\mathrm{x}+1}{({\mathrm{x}}^2+1)({\mathrm{x}}^2+\mathrm{x}+1)}\mathrm{dx}$$$$\mathrm{I}=\frac{\mathrm{x}-1}{{\mathrm{x}}^2+\mathrm{x}+1}+\int \frac{\mathrm{dx}}{{\mathrm{x}}^2+1}=\frac{\mathrm{x}-1}{{\mathrm{x}}^2+\mathrm{x}+1}+\arctan \left(\mathrm{x}\right)+\mathrm{C}$$

Commented by liberty last updated on 13/Feb/21

$$$$$$\mathrm{sch}\ddot{\mathrm{o}ne}\mathrm{L}\ddot{\mathrm{o}sung}$$

Commented by SLVR last updated on 13/Feb/21

$$greatjob...nicesir$$

Commented by SLVR last updated on 13/Feb/21

$$butondifferentiaionofRHS1stpartdenominator$$$$hasnosqare...how...toproceed..$$

Commented by EDWIN88 last updated on 13/Feb/21

$$\mathrm{oh}\mathrm{it}\mathrm{is}\mathrm{only}\mathrm{typo}$$

Commented by SLVR last updated on 13/Feb/21

$$welcomesir..$$

Answered by EDWIN88 last updated on 13/Feb/21

$$\int \frac{\mathrm{P}\left(\mathrm{x}\right)}{\mathrm{Q}\left(\mathrm{x}\right)}\mathrm{dx}=\frac{{\mathrm{P}}_1\left(\mathrm{x}\right)}{{\mathrm{Q}}_1\left(\mathrm{x}\right)}+\int \frac{{\mathrm{P}}_2\left(\mathrm{x}\right)}{{\mathrm{Q}}_2\left(\mathrm{x}\right)}\mathrm{dx}$$$$\mathrm{where}{\mathrm{Q}}_1\left(\mathrm{x}\right)\mathrm{is}\gcd \mathrm{of}\mathrm{Q}\left(\mathrm{x}\right)\mathrm{and}{\mathrm{Q}}^{\prime }\left(\mathrm{x}\right)$$$${\mathrm{Q}}_2\left(\mathrm{x}\right)=\frac{\mathrm{Q}\left(\mathrm{x}\right)}{{\mathrm{Q}}_1\left(\mathrm{x}\right)}\mathrm{and}{\mathrm{P}}_1\left(\mathrm{x}\right)\mathrm{and}{\mathrm{P}}_2\left(\mathrm{x}\right)\mathrm{are}$$$$\mathrm{unknown}\mathrm{polynomials}\mathrm{of}\mathrm{degree}\mathrm{one}\mathrm{less}$$$$\mathrm{than}{\mathrm{Q}}_1\left(\mathrm{x}\right)\mathrm{and}{\mathrm{Q}}_2\left(\mathrm{x}\right)\mathrm{respectively}.$$$$\mathrm{from}\mathrm{the}\mathrm{equation}\mathrm{Q}\left(\mathrm{x}\right)=({\mathrm{x}}^2+1){({\mathrm{x}}^2+\mathrm{x}+1)}^2$$$${\mathrm{Q}}^{\prime }\left(\mathrm{x}\right)=2\mathrm{x}{({\mathrm{x}}^2+\mathrm{x}+1)}^2+2({\mathrm{x}}^2+1)(2\mathrm{x}+1)({\mathrm{x}}^2+\mathrm{x}+1)$$$$\mathrm{the}\gcd \mathrm{of}\mathrm{Q}\left(\mathrm{x}\right)\mathrm{and}{\mathrm{Q}}^{\prime }\left(\mathrm{x}\right)\mathrm{is}{\mathrm{Q}}_1\left(\mathrm{x}\right)={\mathrm{x}}^2+\mathrm{x}+1$$$$\mathrm{and}{\mathrm{Q}}_2\left(\mathrm{x}\right)=\frac{\mathrm{Q}\left(\mathrm{x}\right)}{{\mathrm{Q}}_1\left(\mathrm{x}\right)}=({\mathrm{x}}^2+1)({\mathrm{x}}^2+\mathrm{x}+1)$$

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