# Integral of 1/[(x^3)+1] from 0 to ∞

I=01x3+1dx=2π31.51.209199576156145

## Partial Fraction Decomposition Method

\begin{aligned} \int_0^{\infty} \dfrac{1}{(x+1)(x^2-x+1)} \,dx \end{aligned}.
\begin{aligned} \int_0^{\infty} \dfrac{1}{3(x+1)} \,dx+
\int_0^{\infty} \dfrac{x-2}{3(x^2-x+1)} \,dx \end{aligned}.
\begin{aligned} \dfrac{1}{3} \int_0^{\infty} \dfrac{1}{x+1} \,dx-\dfrac{1}{3} \int_0^{\infty} \dfrac{x-2}{x^2-x+1} \,dx \end{aligned}.
\begin{aligned} \left[\dfrac{1}{3} \ln{(x+1)}\right]\mid_0^{\infty}-\dfrac{1}{6}\int_0^{\infty} \dfrac{2x-1}{x^2-x+1} \,dx+\dfrac{1}{2}\int_0^{\infty} \dfrac{1}{x^2-x+1} \,dx \end{aligned}.
\begin{aligned} \left[\dfrac{1}{3} \ln{(x+1)}-\dfrac{1}{6} \ln{(x^2-x+1)}\right] \mid_0^{\infty}+\dfrac{1}{2}\int_0^{\infty} \dfrac{1}{x^2-x+1} \,dx \end{aligned}
\begin{aligned} \left[\dfrac{1}{3} \ln{(x+1)}-\dfrac{1}{6} \ln{(x^2-x+1)}\right] \mid_0^{\infty}+\dfrac{1}{2}\int_0^{\infty} \dfrac{1}{(x-\dfrac{1}{2})^2+\dfrac{3}{4}} \,dx \end{aligned}
\begin{aligned} \left[\dfrac{1}{3} \ln{(x+1)}-\dfrac{1}{6} \ln{(x^2-x+1)}+\dfrac{1}{\sqrt{3}}\arctan(\dfrac{2x-1}{\sqrt{3}})\right] \mid_0^\infty \end{aligned}
\begin{aligned} \left[\dfrac{1}{6} [2\ln{(x+1)}-\ln{(x^2-x+1)}]+\dfrac{1}{\sqrt{3}}\arctan(\dfrac{2x-1}{\sqrt{3}})\right] \mid_0^\infty \end{aligned}
\begin{aligned} \left[\dfrac{1}{6} \left[\ln\left(\dfrac{{(x+1)^2}}{{(x^2-x+1)}}\right)\right]+\dfrac{1}{\sqrt{3}}\arctan(\dfrac{2x-1}{\sqrt{3}})\right] \mid_0^\infty \end{aligned}

\begin{aligned} \Box \: \: I=\dfrac{2\pi}{3^{1.5}} \approx {1.209199576156145} \end{aligned}

This site uses Akismet to reduce spam. Learn how your comment data is processed.