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

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