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

$$\int\frac{x^4}{x^2-1}dx=\frac{x^3+3x}{3}+\frac{1}{2}\ln\Big|\frac{x-1}{x+1}\Big|+C$$
$$I=\int\frac{x^4}{x^2-1}dx$$ 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: $$I=\int(x^2+1)dx-\frac{1}{2}\int\frac{1}{x+1}dx+\frac{1}{2}\int\frac{1}{x-1}dx$$ $$I=\frac{x^3}{3}+x-\frac{1}{2}\ln|x+1|+\frac{1}{2}\ln|x-1|+C$$ $$I=\frac{x^3+3x}{3}+\frac{1}{2}\ln\Big|\frac{x-1}{x+1}\Big|+C$$