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Question-163587




Question Number 163587 by HongKing last updated on 08/Jan/22
Answered by MJS_new last updated on 08/Jan/22
∫(x/((x+1)(x^2 +1)(a^2 x^2 +1)))dx=  =−(1/(2(a^2 +1)))∫(dx/(x+1))−(1/(2(a^2 −1)))∫((x+1)/(x^2 +1))dx+(a^2 /(a^4 −1))∫(dx/(a^2 x^2 +1))=  =−(1/(2(a^2 +1)))ln ∣x+1∣ −(1/(4(a^2 −1)))ln (x^2 +1) −(1/(2(a^2 −1)))arctan x +(a^2 /(2(a^4 −1)))ln (a^2 x^2 +1) +(a/(a^4 −1))arctan (ax) +C  a>0: Ω(a)=(a^2 /((a^4 −1)))ln a −(((a−1)π)/(4(a+1)(a^2 +1))) and lim Ω(a) =(1/4)_(a→1)   a≤0: Ω(a)∉R
$$\int\frac{{x}}{\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\int\frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{{a}^{\mathrm{2}} }{{a}^{\mathrm{4}} −\mathrm{1}}\int\frac{{dx}}{{a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{ln}\:\mid{x}+\mathrm{1}\mid\:−\frac{\mathrm{1}}{\mathrm{4}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{2}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\mathrm{arctan}\:{x}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}\left({a}^{\mathrm{4}} −\mathrm{1}\right)}\mathrm{ln}\:\left({a}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}\right)\:+\frac{{a}}{{a}^{\mathrm{4}} −\mathrm{1}}\mathrm{arctan}\:\left({ax}\right)\:+{C} \\ $$$${a}>\mathrm{0}:\:\Omega\left({a}\right)=\frac{{a}^{\mathrm{2}} }{\left({a}^{\mathrm{4}} −\mathrm{1}\right)}\mathrm{ln}\:{a}\:−\frac{\left({a}−\mathrm{1}\right)\pi}{\mathrm{4}\left({a}+\mathrm{1}\right)\left({a}^{\mathrm{2}} +\mathrm{1}\right)}\:\underset{{a}\rightarrow\mathrm{1}} {\mathrm{and}\:\mathrm{lim}\:\Omega\left({a}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${a}\leqslant\mathrm{0}:\:\Omega\left({a}\right)\notin\mathbb{R} \\ $$
Commented by HongKing last updated on 08/Jan/22
thank you so much my dear Sir cool
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{cool} \\ $$

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