Question

Solve: $\int \frac{xdx}{x^3+1}$

Answer 1

$\begin{array}{c}I=\int \frac{xdx}{x^3+1}\\ \frac{x}{x^3+1}=\frac{x}{\left(x+1\right)\left(x^2-x+1\right)}=\frac{x+1}{3\left(x^2-x+1\right)}-\frac{1}{3\left(x+1\right)}\\ I=\int \frac{x+1}{3\left(x^2-x+1\right)}dx-\int \frac{1}{3\left(x+1\right)}dx\\ I=\frac{1}{3}\underset{I_1}{\underset{}{\int \frac{x+1}{\left(x^2-x+1\right)}dx}}-\frac{1}{3}\ln \left|x+1\right|\\ in{I}_1\\ x+1=A\frac{d}{dx}\left(x^2-x+1\right)+B\\ \Rightarrow x+1=A\left(2x-1\right)+B\\ 2A=1\Rightarrow A=\frac{1}{2}\\ 1=-A+B\Rightarrow 1=-\frac{1}{2}+B\Rightarrow B=\frac{3}{2}\\ I=\frac{1}{3}\left[\int \frac{\frac{1}{2}\left(2x-1\right)}{\left(x^2-x+1\right)}dx+\frac{3}{2}\int \frac{dx}{\left(x^2-x+1\right)}\right]-\frac{1}{3}\ln \left|x+1\right|\\ I=\frac{1}{6}\int \frac{\left(2x-1\right)}{\left(x^2-x+1\right)}dx+\frac{1}{2}\int \frac{dx}{\left(x^2-x+1\right)}-\frac{1}{3}\ln \left|x+1\right|\\ put,t=x^2-x+1\\ dt=(2x-1)dx\\ I=\frac{1}{6}\int \frac{dt}{t}dx+\frac{1}{2}\int \frac{dx}{x^2-2\times \frac{1}{2}\times x+\frac{1}{4}-\frac{1}{4}+1}-\frac{1}{3}\ln \left|x+1\right|\\ I=\frac{1}{6}\ln \left|t\right|+\frac{1}{2}\int \frac{dx}{{\left(x-\frac{1}{2}\right)}^2+\frac{3}{4}}-\frac{1}{3}\ln \left|x+1\right|\\ I=\frac{1}{6}\ln \left|x^2-x+1\right|+\frac{1}{2}\times \frac{1}{\sqrt{3}/2}{\tan }^{-1}\left(\frac{x-1/2}{\sqrt{3}/2}\right)-\frac{1}{3}\ln \left|x+1\right|\\ I=\frac{1}{6}\ln \left|x^2-x+1\right|+\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{2x-1}{\sqrt{3}}\right)-\frac{1}{3}\ln \left|x+1\right|+C\end{array}$