Some Indefinite Integrals

[jacks]

1. \(\displaystyle \int\frac{x}{1+x^5}dx\)
2. \(\displaystyle \int\frac{x^6}{1+x^5}dx\)

[Grigorios Kostakos]

1. \begin{align*}
\qquad J&=\displaystyle \int\frac{x}{1+x^5}\,dx\\
&=\frac{1}{5}\int{\frac{1+\sqrt {5}+\bigl({1-\sqrt{5}\,}\bigr)\,x}{2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\,dx}+\frac{1}{5}\int{\frac {1-\sqrt{5}+\bigl({1+\sqrt{5}\,}\bigr)\,x}{2x^2+\bigl({-1+\sqrt{5}\,}\bigr)\,x+2}\,dx}-\frac{1}{5}
\int{\frac {1}{1+x}\,dx}\\
&=\frac{1}{5}\,J_1+\frac{1}{5}\,J_2-\frac{1}{5}\,\log|{1+x}|+c_1\,.
\end{align*} \begin{align*}
J_1&=\displaystyle\int{\frac{1+\sqrt{5}+\bigl({1-\sqrt{5}\,}\bigr)\,x}{2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\,dx}\\
&=\frac{1-\sqrt{5}}{4}\int{\frac { 4x-1-\sqrt{5}}{2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\,dx}+\int{\frac {\sqrt{5}}{2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\,dx}\\
&=\frac{1-\sqrt{5}}{4}\,\log\bigl|{2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\bigr|+c_2+\int{\frac{4\sqrt{5}}{\bigl({\frac{4x-1-\sqrt{5}}{\sqrt{2}}\,}\bigr)^2+5-\sqrt{5}}\,dx}\\
&=\frac{1-\sqrt{5}}{4}\,\log\bigl({2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\bigr)+c_2+\frac{\sqrt{2}}{\sqrt{5}-1}\int{\frac{1}{\Bigl({\frac{4x-1-\sqrt{5}}{\sqrt{2}\sqrt{5-\sqrt{5}}}}\Bigr)^2+1}\,d\Bigl({\tfrac{4x-1-\sqrt{5}}{\sqrt{2}\sqrt{5-\sqrt{5}}}}\Bigr)}\\
&=\frac{1-\sqrt{5}}{4}\,\log\bigl({2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\bigr)+\frac{\sqrt{2}}{\sqrt{\sqrt{5}-1}}\,\arctan\Bigl({\tfrac{4x-1-\sqrt{5}}{\sqrt{2}\sqrt{5-\sqrt{5}}}}\Bigr)+c_3\,.
\end{align*} Similarly \[J_2=\displaystyle\frac{1+\sqrt{5}}{4}\,\log\bigl({2x^2-\bigl({1-\sqrt{5}\,}\bigr)\,x+2}\bigr)-\frac{\sqrt{2}}{\sqrt{\sqrt{5}+1}}\,\arctan\Bigl({\tfrac{4x-1+\sqrt{5}}{\sqrt{2}\sqrt{5+\sqrt{5}}}}\Bigr)+c_4\,.\] So \begin{align*}
J&=\frac{1}{5}\,J_1+\frac{1}{5}\,J_2-\frac{1}{5}\,\log|{1+x}|+c_1\\
&=\frac{1-\sqrt{5}}{20}\,\log\bigl({2x^2-\bigl({1+\sqrt{5}\,}\bigr)\,x+2}\bigr)+\frac{\sqrt{2}}{5\sqrt{\sqrt{5}-1}}\,\arctan\Bigl({\tfrac{4x-1-\sqrt{5}}{\sqrt{2}\sqrt{5-\sqrt{5}}}}\Bigr)\,+\\
&\qquad\frac{1+\sqrt{5}}{20}\,\log\bigl({2x^2-\bigl({1-\sqrt{5}\,}\bigr)\,x+2}\bigr)-\frac{\sqrt{2}}{5\sqrt{\sqrt{5}+1}}\,\arctan\Bigl({\tfrac{4x-1+\sqrt{5}}{\sqrt{2}\sqrt{5+\sqrt{5}}}}\Bigr)-\frac{1}{5}\,\log|{1+x}|+c\,.
\end{align*}

[Grigorios Kostakos]

2. \begin{align*}
K&=\displaystyle \int\frac{x^6}{1+x^5}\,dx\quad\Rightarrow\\
K+J&=\displaystyle \int\frac{x^6+x}{1+x^5}\,dx= \int{x\,dx}=\frac{x^2}{2}\quad\Rightarrow\\
K&=\frac{x^2}{2}-J\,.
\end{align*}

[jacks]

Thanks for Nice solution Grigorios Kostakos.