Question

Prove that $\int \frac{x^4(x^{10}-1)}{(x^{20}+3x^{10}+1)}dx=\frac{2}{5}{\tan }^{-1}(x^5+\frac{1}{x^5})$.

Answer 1

Let $I=\int \frac{x^4(x^{10}-1)}{x^{20}+3x^{10}+1}dx$
Put $t=x^5\Rightarrow dt=5x^4$
$I=\frac{1}{5}\int \frac{t^2-1}{t^4+3t^2+1}dt=\frac{1}{5}\int \frac{t^2-1}{(t^2-\frac{\sqrt{5}}{2}+\frac{3}{2})(t^2+\frac{\sqrt{5}}{2}+\frac{3}{2})}dt$
$=\frac{1}{5}\int (\frac{5+\sqrt{5}}{\sqrt{5}(2t^2+\sqrt{5}+3)}+\frac{5-\sqrt{5}}{\sqrt{5}(-2t^2+\sqrt{5}-3)})dt$
$=(\frac{1}{5}+\frac{1}{\sqrt{5}})\int \frac{1}{2t^2+\sqrt{5}+3}dt+(\frac{1}{\sqrt{5}}-\frac{1}{5})\int \frac{1}{-2u^2+\sqrt{5}-3}dt$
$=(\frac{1}{5(3+\sqrt{5})}+\frac{1}{\sqrt{5}(3+\sqrt{5})})\int \frac{1}{\frac{2t^2}{(3+\sqrt{5})}+1}dt$
$+(\frac{1}{\sqrt{5}(\sqrt{5}-3)}-\frac{1}{5(\sqrt{5}-3)})\int \frac{1}{1-\frac{2t^2}{(\sqrt{5}-3)}}dt$
$=-\frac{(\sqrt{5}-1)\sqrt{\frac{2}{3+\sqrt{5}}}({\tan }^{-1}\left(\sqrt{\frac{2}{3+\sqrt{5}}}{x}^5\right)-{\tan }^{-1}\left(\sqrt{\frac{1}{2}(3+\sqrt{5})}{x}^5\right))}{5(\sqrt{5}-3)}$
$=-\frac{1}{5}{\tan }^{-1}\left(\frac{x^5}{x^{10}+1}\right)$