Question

Evaluate the following integral:

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

Answer 1

\[\text{Let }I = \int\frac{x^3 + x + 1}{x^2 - 1}dx\]

 

\[\text{Here the integrand }\frac{x^3 + x + 1}{x^2 - 1}\text{ is not a proper rational function, so we divide }x^3 + x + 1\text{ by }x^2 - 1\text{ and find that}\]

 

\[\frac{x^3 + x + 1}{x^2 - 1} = x + \frac{2x + 1}{x^2 - 1} = x + \frac{2x + 1}{\left( x + 1 \right)\left( x - 1 \right)}\]

 

\[\text{Let }\frac{2x + 1}{\left( x + 1 \right)\left( x - 1 \right)} = \frac{A}{x + 1} + \frac{B}{x - 1}\]

 

\[\Rightarrow 2x + 1 = A\left( x - 1 \right) + B\left( x + 1 \right)\]

 

Equating the coefficients of x and constants, we get

 

\[2 = A + B\text{ and }1 = - A + B\]

 

\[\text{or }A = \frac{1}{2}\text{ and }B = \frac{3}{2}\]

 

\[\therefore I = \int\left( x + \frac{\frac{1}{2}}{x + 1} + \frac{\frac{3}{2}}{x - 1} \right)dx\]

 

\[= \int x\ dx + \frac{1}{2}\int\frac{1}{x + 1}dx + \frac{3}{2}\int\frac{1}{x - 1} dx\]

 

\[= \frac{x^2}{2} + \frac{1}{2}\log\left| x + 1 \right| + \frac{3}{2}\log\left| x - 1 \right| + c\]

 

\[= \frac{x^2}{2} + \frac{1}{2}\log\left| x + 1 \right| + \frac{3}{2}\log\left| x - 1 \right| + c\]

 

\[\text{Hence, }\int\frac{x^3 + x + 1}{x^2 - 1}dx = \frac{x^2}{2} + \frac{1}{2}\log\left| x + 1 \right| + \frac{3}{2}\log\left| x - 1 \right| + c\]