x-x-2-a-2-x-3-b-2-

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

Ostrogradsky . . method .

Commented by MJS_new last updated on 17/Nov/20

you should search this method on the www

to learn when {it}^{'} s possible to use it . here {it}^{'} s

not possible .

Answered by MJS_new last updated on 17/Nov/20

this is the solution of \int \frac{d x}{(x^{2} + a^{2}) (x^{3} + b^{2})}

we must decompose \dots I 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}} d x

I_{2} = \frac{1}{3 (a^{2} + b^{4 / 3}) b^{4 / 3}} \int \frac{d x}{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{2 x - b^{2 / 3}}{x^{2} - b^{2 / 3} x + b^{4 / 3}} d x

I_{4} = \frac{a^{2} - b^{4 / 3}}{2 (a^{4} - a^{2} b^{4 / 3} + b^{8 / 3}) b^{2 / 3}} \int \frac{d x}{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 ∣ 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}} \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{(2 x - b^{2 / 3}) \sqrt{3}}{3 b^{2 / 3}}

Commented by ajfour last updated on 17/Nov/20

o o p s!

Commented by MJS_new last updated on 17/Nov/20

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

Commented by benjo_mathlover last updated on 17/Nov/20

w a w \dots .

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}} d x

J_{2} = - \frac{a^{4}}{a^{6} + b^{4}} \int \frac{d x}{x^{2} + a^{2}}

J_{3} = - \frac{1}{3 (a^{2} + b^{4 / 3}) b^{2 / 3}} \int \frac{d x}{x + b^{2 / 3}}

J_{4} = \frac{a^{2} - 2 b^{4 / 3}}{6 (a^{4} - a^{2} b^{4 / 3} + b^{8 / 3}) b^{2 / 3}} \int \frac{2 x - 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{d x}{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 ∣ x + b^{2 / 3} ∣

J_{4} = \frac{a^{2} - 2 b^{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{(2 x - b^{2 / 3}) \sqrt{3}}{3 b^{2 / 3}}

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