Question

Evaluate integral of x³-1/x³​+x dx

Answer 1

$$\begin{gathered} \int \frac{x^3-1}{x^3+x}dx \\ =\int \left(1-\frac{x+1}{x^3+x}\right)dx \\ =\int dx-\int \frac{x+1}{x^3+x}dx \\ =\int dx-\int \frac{x+1}{x\left(x^2+1\right)}dx \\ =\int dx-\int \frac{1}{x^2+1}dx\mathbf{-}\int \frac{1}{x\left(x^2+1\right)}dx \\ =\int dx-\int \frac{1}{x^2+1}dx\mathbf{-}\int \frac{1}{x^3\left(1+{\displaystyle \frac{1}{x^2}}\right)}dx \\ =\int dx-\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{dz}{z},\mathrm{where}1+\frac{1}{x^2}=z\Rightarrow -\frac{dz}{2}=\frac{dx}{x^3} \\ =x-{\tan }^{-1}x+\frac{1}{2}\log z+C \\ =x-{\tan }^{-1}x+\frac{1}{2}\log \left(1+\frac{1}{x^2}\right)+C \end{gathered}$$