Question

Integrate the rational functions.
$\int \frac{1}{x^4-1}dx$

Answer 1

$\int \frac{1}{x^4-1}dx=\int \frac{1}{(x^2-1)(x^2+1)}dx[\because a^2-b^2=(a+b\left)\right(a-b\left)\right]$
Let $\frac{1}{(x^2-1)(x^2+1)}=\frac{Ax+B}{(x^2+1)}+\frac{Cx+D}{(x^2-1)}$
$\Rightarrow \frac{1}{(x^2+1)(x^2+1)}=\frac{(Ax+B)(x^2-1)+(Cx+D)(x^2+1)}{(x^2+1)(x^2-1)}\Rightarrow 1=A{x}^3+B{x}^2-Ax-B+C{x}^3+x^{2}D+Cx+D\Rightarrow 1=x^3(A+C)+x^2(B+D)+x(-A+C)+(-B+D)$
On comparing the coefficients of $x^3,x^2$, x and constant term on both sides, we get
A+C=0 ...(i)
B+D =0 ...(ii)
-A+C =0 ...(iii)
and -B+D =1 ....(iv)
On adding Eqs. (i) and (iii), we get
$2C=0\Rightarrow C=0\Rightarrow C=0$, then $A=0$ [from Eq.(i)]
On adding Eqs.(ii)and (iv),we get $2D=1\Rightarrow D=\frac{1}{2}$
On putting the value of D in Eq. (iv), we get
$-B+\frac{1}{2}=1\Rightarrow B=-\frac{1}{2}\therefore A=0,B=-\frac{1}{2},C=0\text{and}D=\frac{1}{2}\therefore \int \frac{1}{(x^4-1)}dx=\int \frac{Ax+B}{(x^2+1)}dx+\int \frac{Cx+D}{(x^2-1)}dx=-\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2-1}dx=-\frac{1}{2}ta{n}^{-1}x+\frac{1}{2}.\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C=-\frac{1}{2}ta{n}^{-1}x+\frac{1}{4}log\left|\frac{x-1}{x+1}\right|+C$
Alternative Method
$\int \frac{1}{(x^2-1)(x^2+1)}dx$

On putting $x^2=t$, we get
$\frac{1}{(t-1)(t+1)}=\frac{A}{(t-1)}+\frac{B}{(t+1)}\text{[by partial fraction]....(i)}\Rightarrow \frac{1}{(t-1)(t+1)}=\frac{A(t+1)+B(t-1)}{(t-1)(t+1)}\Rightarrow 1=At+A+Bt-B\Rightarrow 1=t(A+B)+(A-B)$
On comparing the coefficients of t and constant term on both sides, we get
A+B =0 ....(ii)
A - B =1 ......(iii)
On adding Eqs.(ii) and (iii), we get
$A=\frac{1}{2},\text{and from Eq.}\left(iii\right),B=-\frac{1}{2}$
On putting the values of A and B in Eq. (i), we get
$\frac{1}{(t-1)(t+1)}=\frac{\frac{1}{2}}{t-1}+\frac{-\frac{1}{2}}{t+1}\Rightarrow \frac{1}{(x^2-1)(x^2+1)}=\frac{1}{2}\times \frac{1}{x^2-1}-\frac{1}{2}\times \frac{1}{x^2-1}\text{[on putting t =}{x}^2\text{]}\therefore \int \frac{dx}{(x^2-1)(x^2+1)}=-\frac{1}{2}\int \frac{1}{x^2+1}dx+\frac{1}{2}\int \frac{1}{x^2-1}dx=-\frac{1}{2}ta{n}^{-1}x+\frac{1}{2}.\frac{1}{2}log\left|\frac{x-1}{x+1}\right|+C=-\frac{1}{2}ta{n}^{-1}x+\frac{1}{4}log\left|\frac{x-1}{x+1}\right|+C$