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Question Number 122461 by benjo_mathlover last updated on 17/Nov/20

$$\int \frac{x}{(x^2+a^2)(x^3+b^2)}?$$

Commented by liberty last updated on 17/Nov/20

$$\mathrm{Ostrogradsky}..\mathrm{method}.$$

Commented by MJS_new last updated on 17/Nov/20

$$\mathrm{you}\mathrm{should}\mathrm{search}\mathrm{this}\mathrm{method}\mathrm{on}\mathrm{the}\mathrm{www}$$$$\mathrm{to}\mathrm{learn}\mathrm{when}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{possible}\mathrm{to}\mathrm{use}\mathrm{it}.\mathrm{here}{\mathrm{it}}^{\prime }\mathrm{s}$$$$\mathrm{not}\mathrm{possible}.$$

Answered by MJS_new last updated on 17/Nov/20

$$\mathrm{this}\mathrm{is}\mathrm{the}\mathrm{solution}\mathrm{of}\int \frac{dx}{(x^2+a^2)(x^3+b^2)}$$$$\mathrm{we}\mathrm{must}\mathrm{decompose}...\mathrm{I}\mathrm{get}$$$$I=I_1+I_2+I_3+I_4+C$$$$I_1=\frac{a^2}{a^6+b^4}\int \frac{x}{x^2+a^2}dx$$$$I_2=\frac{1}{3(a^2+b^{4/3})b^{4/3}}\int \frac{dx}{x+b^{2/3}}$$$$I_3=-\frac{a^2+b^{4/3}}{6(a^4-a^2{b}^{4/3}+b^{8/3})b^{4/3}}\int \frac{2x-b^{2/3}}{x^2-b^{2/3}x+b^{4/3}}dx$$$$I_4=\frac{a^2-b^{4/3}}{2(a^4-a^2{b}^{4/3}+b^{8/3})b^{2/3}}\int \frac{dx}{x^2-b^{2/3}x+b^{4/3}}$$$$$$$$I_1=\frac{a^2}{2(a^6+b^4)}\ln (x^2+a^2)$$$$I_2=\frac{1}{3(a^2+b^{4/3})b^{4/3}}\ln \mid x+b^{2/3}\mid$$$$I_3=-\frac{a^2+b^{4/3}}{6(a^4-a^2{b}^{4/3}+b^{8/3})b^{4/3}}\ln (x^2-b^{2/3}x+b^{4/3})$$$$I_4=\frac{(a^2-b^{4/3})\sqrt{3}}{3(a^4-a^2{b}^{4/3}+b^{8/3})b^{4/3}}\arctan \frac{(2x-b^{2/3})\sqrt{3}}{3b^{2/3}}$$

Commented by ajfour last updated on 17/Nov/20

$$oops!$$

Commented by MJS_new last updated on 17/Nov/20

$$\mathrm{sorry}\mathrm{wrong}.\mathrm{I}\mathrm{solved}\int \frac{dx}{(x^2+a^2)(x^3+b^2)}$$

Commented by benjo_mathlover last updated on 17/Nov/20

$$waw....$$

Answered by MJS_new last updated on 17/Nov/20

$$\int \frac{x}{(x^2+a^2)(x^3+b^2)}=J_1+J_2+J_3+J_4+J_5+C$$$$$$$$J_1=\frac{b^2}{a^6+b^4}\int \frac{x}{x^2+a^2}dx$$$$J_2=-\frac{a^4}{a^6+b^4}\int \frac{dx}{x^2+a^2}$$$$J_3=-\frac{1}{3(a^2+b^{4/3})b^{2/3}}\int \frac{dx}{x+b^{2/3}}$$$$J_4=\frac{a^2-2b^{4/3}}{6(a^4-a^2{b}^{4/3}+b^{8/3})b^{2/3}}\int \frac{2x-b^{2/3}}{x^2-b^{2/3}x+b^{4/3}}$$$$J_5=\frac{a^2}{2(a^4-a^2{b}^{4/3}+b^{8/3})}\int \frac{dx}{x^2-b^{2/3}x+b^{4/3}}$$$$$$$$J_1=\frac{b^2}{2(a^6+b^4)}\ln (x^2+a^2)$$$$J_2=-\frac{a^3}{a^6+b^4}\arctan \frac{x}{a}$$$$J_3=-\frac{1}{3(a^2+b^{4/3})b^{2/3}}\ln \mid x+b^{2/3}\mid$$$$J_4=\frac{a^2-2b^{4/3}}{6(a^4-a^2{b}^{4/3}+b^{8/3})b^{2/3}}\ln (x^2-b^{2/3}x+b^{4/3})$$$$J_5=\frac{a^2\sqrt{3}}{3(a^4-a^2{b}^{4/3}+b^{8/3})b^{2/3}}\arctan \frac{(2x-b^{2/3})\sqrt{3}}{3b^{2/3}}$$

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