dx-x-3-x-2-2x-5-4-

Question Number 77817 by Dah Solu Tion last updated on 10/Jan/20

$$\int\frac{{dx}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{5}\right)^{\mathrm{4}} } \\ $$$$ \\ $$

Commented by mathmax by abdo last updated on 10/Jan/20

we decompose F(x)=(1/(x^3 (x^2 +2x+5)^4 )) inside C(x) x^(2 ) +2x +5 =0 →Δ^′ =1−5=−4 ⇒x_1 =−1+2i and x_2 =−1−2i x_1 =(√5)e^(iarctan(−2)) =(√5)e^(−iarctan(2)) and x_2 =(√5)e^(i arctan(2)) ⇒ F(x)=(1/(x^3 (x−(√5)e^(iarctan(2)) )^4 (x−(√5)e^(−iarctan(2)) )^4 )) =(a/x) +(b/x^2 ) +(c/x^3 ) +Σ_(i=1) ^4 (a_i /((x−(√5)e^(iarctan(2)) )^i )) +Σ_(i=1) ^4 (b_i /((x−(√5)e^(−iarctan(2)) )^i )) ⇒∫ F(x)dx =aln∣x∣−(b/x) −(c/(2x^2 )) +Σ_(i=1) ^4 a_i ∫ (dx/((x−(√5)e^(iarctan(2)) )^i )) +Σ_(i=1) ^4 b_i ∫ (dx/((x−(√5)e^(−i arctan(2)) )^i )) +C rest to find a_i and b_i ....be continued ...

$${we}\:{decompose}\:{F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}^{\mathrm{2}} \:+\mathrm{2}{x}+\mathrm{5}\right)^{\mathrm{4}} }\:{inside}\:{C}\left({x}\right) \\ $$$${x}^{\mathrm{2}\:} +\mathrm{2}{x}\:+\mathrm{5}\:=\mathrm{0}\:\rightarrow\Delta^{'} =\mathrm{1}−\mathrm{5}=−\mathrm{4}\:\Rightarrow{x}_{\mathrm{1}} =−\mathrm{1}+\mathrm{2}{i}\:{and}\:{x}_{\mathrm{2}} =−\mathrm{1}−\mathrm{2}{i} \\ $$$${x}_{\mathrm{1}} =\sqrt{\mathrm{5}}{e}^{{iarctan}\left(−\mathrm{2}\right)} \:=\sqrt{\mathrm{5}}{e}^{−{iarctan}\left(\mathrm{2}\right)} \:{and}\:{x}_{\mathrm{2}} =\sqrt{\mathrm{5}}{e}^{{i}\:{arctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{3}} \left({x}−\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\mathrm{2}\right)} \right)^{\mathrm{4}} \left({x}−\sqrt{\mathrm{5}}{e}^{−{iarctan}\left(\mathrm{2}\right)} \right)^{\mathrm{4}} } \\ $$$$=\frac{{a}}{{x}}\:+\frac{{b}}{{x}^{\mathrm{2}} }\:+\frac{{c}}{{x}^{\mathrm{3}} }\:+\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} \:\frac{{a}_{{i}} }{\left({x}−\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\mathrm{2}\right)} \right)^{{i}} }\:+\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} \:\frac{{b}_{{i}} }{\left({x}−\sqrt{\mathrm{5}}{e}^{−{iarctan}\left(\mathrm{2}\right)} \right)^{{i}} } \\ $$$$\Rightarrow\int\:{F}\left({x}\right){dx}\:={aln}\mid{x}\mid−\frac{{b}}{{x}}\:−\frac{{c}}{\mathrm{2}{x}^{\mathrm{2}} }\:+\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} {a}_{{i}} \:\int\:\:\frac{{dx}}{\left({x}−\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\mathrm{2}\right)} \right)^{{i}} } \\ $$$$+\sum_{{i}=\mathrm{1}} ^{\mathrm{4}} \:{b}_{{i}} \int\:\:\:\:\:\frac{{dx}}{\left({x}−\sqrt{\mathrm{5}}{e}^{−{i}\:{arctan}\left(\mathrm{2}\right)} \right)^{{i}} }\:+{C}\: \\ $$$${rest}\:{to}\:{find}\:{a}_{{i}} \:{and}\:{b}_{{i}} ….{be}\:{continued}\:… \\ $$

Commented by MJS last updated on 11/Jan/20

look up Ostrogradski′s method on the web; I think it′s the best in cases like this one. I have no time now, will post the solution later

$$\mathrm{look}\:\mathrm{up}\:\mathrm{Ostrogradski}'\mathrm{s}\:\mathrm{method}\:\mathrm{on}\:\mathrm{the}\:\mathrm{web}; \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{best}\:\mathrm{in}\:\mathrm{cases}\:\mathrm{like}\:\mathrm{this}\:\mathrm{one}.\:\mathrm{I}\:\mathrm{have} \\ $$$$\mathrm{no}\:\mathrm{time}\:\mathrm{now},\:\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{solution}\:\mathrm{later} \\ $$

Commented by Dah Solu Tion last updated on 11/Jan/20

$${Thanks}\:{man} \\ $$$${i}\:{will}\:{try}\:{d}\:{rest} \\ $$

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