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# Integrate the rational functions$\frac{x^3+x+1}{x^2-1}$

Toolbox:
• (i)If the rational is improper in nature we can divide and separate the terms to make it a proper rational function.
• (ii) $\frac{Px+q}{(x+a)(x+b)}=\frac{A}{(x+a)}+\frac{B}{(x+b)}$
• (iii)$\int \frac{dx}{x+a}=log| x+a |+c$
Given$I=\int \frac{x^3+x+1}{x^2-1}$

Since it is an improper rational function, let us divide

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

$=x+\frac{2x+1}{(x+1)(x-1)}$

$\frac{2x+1}{(x+1)(x-1)}=\frac{A}{(x+1)}+\frac{B}{(x-1)}$

2x+1=A(x-1)+B(x+1)

Equating the coefficients of x,

2=A+B ------(1)

Equating the constant terms,

1=-A+B -----(2)

Now Add equation (1) and equation(2)

A+B=2
-A+B=1
_________________
2B=3=>B=3/2
Substitute for B in equ (1)

A+B=2

$A+\frac{3}{2}=2$

A=1/2

Hence A=1/2 and B=3/2

Now substitude for A and B

$\frac{2x+1}{(x+1)(x-1)}=\frac{1}{2(x+1)}+\frac{3}{2(x-1)}$

On integrating we get,

Therefore $I=\int xdx+\frac{1}{2}\int \frac {dx}{x+1}+\frac{3}{2}\int \frac{dx}{(x-1)}$

$\frac{x^2}{2}+\frac{1}{2}log |x+1|+\frac{3}{2} log |x-1|+c$