Let $I = \int \frac{x^3+x+1}{x^2-1} dx$ Since $\frac{x^3+x+1}{x^2-1}$ is an improper fraction, therefore, we convert it into a proper fraction by long division method. $\frac{x^3+x+1}{x^2-1} = x + \frac{2x+1}{x^2-1}$ ...(i) Now, $\frac{2x+1}{x^2-1}$ $= \frac{2x+1}{(x+1)(x-1)}$ $= \frac{A}{x+1} + \frac{B}{x-1}$⇒ 2x + 1 = A(x – 1) + B(x + 1) ...(ii) Putting x = – 1 in (ii), we get –2 + 1 = A(–1 – 1) ⇒ A $= \frac{-1}{-2} = \frac{1}{2}$ Putting x = 1 in (ii), we get 2 + 1 = B(1 + 1) ⇒ B $= \frac{3}{2}$ $\therefore \frac{2x+1}{x^2-1} = \frac{1}{2(x+1)} + \frac{3}{2(x-1)}$ ...(iii) From (i) and (iii), $\frac{x^3+x+1}{x^2-1}$ $= x + \frac{1}{2(x+1)} + \frac{3}{2(x-1)}$$\therefore \int \frac{x^3+x+1}{x^2-1} dx$ $= \int x dx + \frac{1}{2} \int \frac{dx}{x+1} + \frac{3}{2} \int \frac{dx}{x-1}$$= \frac{x^2}{2} + \frac{1}{2} \log|x+1| + \frac{3}{2} \log|x-1| + C$