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

$$\int\frac{x^2+x}{x^4-3x^2-4}dx=\frac{1}{10}\ln\frac{|x-2|^3}{(x^2+1)|x+2|}+\frac{1}{5}\tan^{-1}x+C$$
$$A=\int\frac{x^2+x}{x^4-3x^2-4}dx=\int\frac{x^2+x}{(x^2+1)(x^2-4)}dx=\int\frac{x^2+x}{(x^2+1)(x-2)(x+2)}dx$$ 1) Express the integrand as a sum of partial fractions: Here we have $x^2+1$ as an irreducible quadratic factor, $$\frac{x^2+x}{(x^2+1)(x-2)(x+2)}dx=\frac{Ax+B}{x^2+1}+\frac{C}{x-2}+\frac{D}{x+2}$$ Clear fractions: $$(Ax+B)(x^2-4)+C(x^2+1)(x+2)+D(x^2+1)(x-2)=x^2+x$$ $$Ax^3+Bx^2-4Ax-4B+Cx^3+2Cx^2+Cx+2C+Dx^3-2Dx^2+Dx-2D=x^2+x$$ $$(A+C+D)x^3+(B+2C-2D)x^2+(-4A+C+D)x+(-4B+2C-2D)=x^2+x$$ Equating coefficients of corresponding powers of $x$, we get $A+C+D=0$ $B+2C-2D=1$ $-4A+C+D=1$ $-4B+2C-2D=0$ Solving the system of equations, we get $A=-1/5$, $B=1/5$, $C=3/10$ and $D=-1/10$ Therefore, $$\frac{x^2+x}{(x^2+1)(x-2)(x+2)}dx=-\frac{1}{5}\frac{x-1}{x^2+1}+\frac{3}{10}\frac{1}{x-2}-\frac{1}{10}\frac{1}{x+2}$$ 2) Evaluate the integral: $$A=-\frac{1}{5}\int\frac{x}{x^2+1}dx+\frac{1}{5}\int\frac{1}{x^2+1}dx+\frac{3}{10}\int\frac{1}{x-2}dx-\frac{1}{10}\int\frac{1}{x+2}dx$$ $$A=-\frac{1}{10}\int\frac{1}{x^2+1}d(x^2+1)+\frac{1}{5}\tan^{-1}x+\frac{3}{10}\ln|x-2|-\frac{1}{10}\ln|x+2|$$ $$A=-\frac{1}{10}\ln|x^2+1|+\frac{1}{5}\tan^{-1}x+\frac{1}{10}\ln|x-2|^3-\frac{1}{10}\ln|x+2|+C$$ $$A=\frac{1}{10}\ln\frac{|x-2|^3}{(x^2+1)|x+2|}+\frac{1}{5}\tan^{-1}x+C$$