How to evaluate the following integral?

Question

How do I evaluate the following expression?

$$\int_1^3\frac{1+x}{1+x^3}~\mathrm{d}x$$

I am trying this question by integrating the expressions separately like $\frac{1}{1+x^3}$ and $\frac{x}{1+x^3}$. But can't solve after that. How to proceed further?

Answer 1

Observe that

\begin{align*} \frac{1+x}{1+x^3} &=\frac{1+x}{(1+x)(x^2-x+1)} \\ &=\frac{1}{x^2-x+1} \\ &=\frac{1}{\left(x-\frac{1}{2}\right)^2-\frac{1}{4}+1} \\ &=\frac{4}{3\left(\left(\frac{2x-1}{\sqrt{3}}\right)^2+1\right)}, \end{align*}

so that we, with the substitution $\xi=\frac{2x-1}{\sqrt{3}}$, $\mathrm{d}\xi=\frac{2}{\sqrt{3}}~\mathrm{d}x$, get that

\begin{align*} \int_1^3\frac{1+x}{1+x^3}~\mathrm{d}x &=\frac{2}{\sqrt{3}}\int_{\frac{1}{\sqrt{3}}}^{\frac{5}{\sqrt{3}}}\frac{\mathrm{d}\xi}{\xi^2+1} \\ &=\frac{2}{\sqrt{3}}\biggl[\arctan\xi\biggr]_{\frac{1}{\sqrt{3}}}^{\frac{5}{\sqrt{3}}} \\ &=\frac{2}{\sqrt{3}}\left(\arctan\frac{5}{\sqrt{3}}-\arctan\frac{1}{\sqrt{3}}\right). \end{align*}