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Question Number 163587 by HongKing last updated on 08/Jan/22

Answered by MJS_new last updated on 08/Jan/22

$$\int \frac{x}{(x+1)(x^2+1)(a^2{x}^2+1)}dx=$$$$=-\frac{1}{2(a^2+1)}\int \frac{dx}{x+1}-\frac{1}{2(a^2-1)}\int \frac{x+1}{x^2+1}dx+\frac{a^2}{a^4-1}\int \frac{dx}{a^2{x}^2+1}=$$$$=-\frac{1}{2(a^2+1)}\ln \mid x+1\mid -\frac{1}{4(a^2-1)}\ln (x^2+1)-\frac{1}{2(a^2-1)}\arctan x+\frac{a^2}{2(a^4-1)}\ln (a^2{x}^2+1)+\frac{a}{a^4-1}\arctan \left(ax\right)+C$$$$a>0:\mathrm{\Omega }\left(a\right)=\frac{a^2}{(a^4-1)}\ln a-\frac{(a-1)\pi }{4(a+1)(a^2+1)}\underset{a\to 1}{\mathrm{and}\lim \mathrm{\Omega }\left(a\right)=\frac{1}{4}}$$$$a⩽0:\mathrm{\Omega }\left(a\right)\notin \mathbb{R}$$

Commented by HongKing last updated on 08/Jan/22

$$\mathrm{thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathrm{my}\mathrm{dear}\mathrm{Sir}\mathrm{cool}$$

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