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Question Number 125482 by joki last updated on 11/Dec/20

Answered by MJS_new last updated on 11/Dec/20

$$\int \frac{4x^3+x^2+1}{{(x^2-2)}^3}dx=$$$$\left[{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}\right]$$$$=\frac{x^3-64x^2-14x+64}{32{(x^2-2)}^2}+\frac{1}{32}\int \frac{dx}{x^2-2}=$$$$=\frac{x^3-64x^2-14x+64}{32{(x^2-2)}^2}+\frac{\sqrt{2}}{128}\ln \mid \frac{\sqrt{2}x-2}{\sqrt{2}x+2}\mid +C$$

Commented by MJS_new last updated on 11/Dec/20

$$\mathrm{please}\mathrm{search}\mathrm{for}{\mathrm{Ostrogradski}}^{\prime }\mathrm{s}\mathrm{Method}$$$$\mathrm{here}\mathrm{or}\mathrm{on}\mathrm{the}\mathrm{www}.{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{understand}.$$$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{have}\mathrm{the}\mathrm{time}\mathrm{now}\mathrm{to}\mathrm{explain}\mathrm{it}\mathrm{again}$$

Commented by joki last updated on 11/Dec/20

$$ldontunderstandsir.pleaseexplainwithdetail.thankssir$$

Commented by MJS_new last updated on 14/Dec/20

$$\int \frac{P\left(x\right)}{Q\left(x\right)}dx=\frac{P_1\left(x\right)}{Q_1\left(x\right)}+\int \frac{P_2\left(x\right)}{Q_2\left(x\right)}dx$$$$\mathrm{where}P\left(x\right),Q\left(x\right)\mathrm{are}\mathrm{polynomes}\mathrm{and}\mathrm{the}$$$$\mathrm{degree}\mathrm{of}P\left(x\right)<\mathrm{the}\mathrm{degree}\mathrm{of}Q\left(x\right)\mathrm{and}\mathrm{the}$$$$\mathrm{fraction}\mathrm{cannot}\mathrm{be}\mathrm{further}\mathrm{simplified}.$$$$\mathrm{then}{Q}_1\left(x\right)=\gcd (Q\left(x\right),Q^{{}^{\prime }}\left(x\right))$$$$\mathrm{and}{Q}_2\left(x\right)=\frac{Q\left(x\right)}{Q_1\left(x\right)}$$$$P_1\left(x\right)\mathrm{and}{P}_2\left(x\right)\mathrm{are}\mathrm{polynomes}\mathrm{with}\mathrm{degrees}$$$$\mathrm{smaller}\mathrm{than}\mathrm{degrees}\mathrm{of}{Q}_1\left(x\right)\mathrm{resp}.Q_2\left(x\right)$$$$\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{constants}\mathrm{of}{P}_1\left(x\right),P_2\left(x\right)\mathrm{by}$$$$\mathrm{comparing}\mathrm{the}\mathrm{constants}\mathrm{of}\mathrm{both}\mathrm{sides}\mathrm{of}$$$$\frac{P\left(x\right)}{Q\left(x\right)}=\frac{d}{dx}\left[\frac{P_1\left(x\right)}{Q_1\left(x\right)}\right]+\frac{P_2\left(x\right)}{Q_2\left(x\right)}$$

Commented by joki last updated on 14/Dec/20

$$\mathrm{can}\mathrm{you}\mathrm{give}\mathrm{me}\mathrm{formula}\mathrm{your}\mathrm{answer}\mathrm{sir}?\mathrm{thanks}\mathrm{sir}$$

Commented by joki last updated on 15/Dec/20

$$\mathrm{l}\mathrm{know}\mathrm{sir}\mathrm{that}\mathrm{formula}{\mathrm{ostrogradsky}}^{\prime }\mathrm{s}\mathrm{method}.$$$$\mathrm{but}\mathrm{i}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}\mathrm{to}\mathrm{formula}{({\mathrm{x}}^2-2)}^3\mathrm{to}\mathrm{found}{\mathrm{P}}_1\mathrm{and}{\mathrm{P}}_2\mathrm{used}$$$${\mathrm{Ax}}^3+{\mathrm{Bx}}^2.....\mathrm{or}\mathrm{to}\mathrm{found}{\mathrm{P}}_1\mathrm{and}{\mathrm{P}}_2\mathrm{used}{\mathrm{Ax}}^2+\mathrm{Bx}+\mathrm{c}$$$$$$

Commented by Tawa11 last updated on 06/Nov/21

$$\mathrm{Great}$$

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