Question

$\int \frac{x^3+x+1}{x^2-1}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \left(\frac{x^3+x+1}{x^2-1}\right)dx \end{gathered}$$

As Degree of Numerator is greater than Degree of Denominator we divide numerator by denominator

$$\begin{gathered} x^2-1\stackrel{x}{\overline{)x^3+x+1}} \\ {x}^3-x \\ \underline{-+} \\ 2x+1 \\ \therefore \frac{x^3+x+1}{x^2-1}=x+\frac{2x+1}{x^2-1}.....\left(1\right) \\ \Rightarrow \frac{x^3+x+1}{x^2-1}=x+\frac{2x}{x^2-1}+\frac{1}{x^2-1} \\ \mathrm{Then}, \\ I=\int xdx+\int \frac{2x}{x^2-1}+\int \frac{dx}{x^2-1^2} \\ \mathrm{Putting}{x}^2-1=t \\ \Rightarrow 2xdx=dt \\ \therefore I=\int xdx+\int \frac{dt}{t}+\int \frac{dx}{x^2-1^2} \\ =\frac{x^2}{2}+\log \left|t\right|+\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C \\ =\frac{x^2}{2}+\log \left|x^2-1\right|+\frac{1}{2}\log \left|\frac{x-1}{x+1}\right|+C \end{gathered}$$