## University Calculus: Early Transcendentals (3rd Edition)

$$\int\frac{x^2}{x^4-1}dx=\frac{1}{4}\ln|x-1|-\frac{1}{4}\ln|x+1|+\frac{1}{2}\tan^{-1}x+C$$
$$A=\int\frac{x^2}{x^4-1}dx=\int\frac{x^2}{(x-1)(x+1)(x^2+1)}dx$$ 1) Express the integrand as a sum of partial fractions: Here we have $x^2+1$ as an irreducible quadratic factor, $$\frac{x^2}{(x-1)(x+1)(x^2+1)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{Cx+D}{x^2+1}$$ Clear fractions: $$A(x+1)(x^2+1)+B(x-1)(x^2+1)+(Cx+D)(x^2-1)=x^2$$ $$Ax^3+Ax^2+Ax+A+Bx^3-Bx^2+Bx-B+Cx^3+Dx^2-Cx-D=x^2$$ $$(A+B+C)x^3+(A-B+D)x^2+(A+B-C)x+(A-B-D)=x^2$$ Equating coefficients of corresponding powers of $x$, we get $A+B+C=0$ $A-B+D=1$ $A+B-C=0$ $A-B-D=0$ Solving the system of equations, we get $A=1/4$, $B=-1/4$, $C=0$ and $D=1/2$ Therefore, $$\frac{x^2}{(x-1)(x+1)(x^2+1)}=\frac{1}{4}\frac{1}{x-1}-\frac{1}{4}\frac{1}{x+1}+\frac{1}{2}\frac{1}{x^2+1}$$ 2) Evaluate the integral: $$A=\frac{1}{4}\int\frac{dx}{x-1}-\frac{1}{4}\int\frac{dx}{x+1}+\frac{1}{2}\int\frac{dx}{x^2+1}$$ $$A=\frac{1}{4}\ln|x-1|-\frac{1}{4}\ln|x+1|+\frac{1}{2}\tan^{-1}x+C$$