Answer
$$\int\frac{x^4}{x^2-1}dx=\frac{x^3+3x}{3}+\frac{1}{2}\ln\Big|\frac{x-1}{x+1}\Big|+C$$
Work Step by Step
$$I=\int\frac{x^4}{x^2-1}dx$$
1) Perform long division:
We perform long division to rewrite the fraction as a polynomial plus a proper fraction, which is $$\frac{x^4}{x^2-1}=x^2+1+\frac{1}{x^2-1}$$
2) Express the remainder fraction as a sum of partial fractions:
$$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}=\frac{A}{x+1}+\frac{B}{x-1}$$
Clear fractions:
$$Ax-A+Bx+B=1$$ $$(A+B)x+(B-A)=1$$
Equating coefficients of corresponding powers of $x$, we get
$A+B=0$
$B-A=1$
Solving for $A$ and $B$, we get $A=-1/2$ and $B=1/2$.
Therefore,
$$\frac{1}{x^2-1}=\frac{1}{(x-1)(x+1)}=-\frac{1}{2}\frac{1}{x+1}+\frac{1}{2}\frac{1}{x-1}$$
2) Evaluate the integral:
$$I=\int(x^2+1)dx-\frac{1}{2}\int\frac{1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$$ $$I=\frac{x^3}{3}+x-\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|+C$$ $$I=\frac{x^3+3x}{3}+\frac{1}{2}\ln\Big|\frac{x-1}{x+1}\Big|+C$$
