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

$$\int^0_{-1}\frac{x^3dx}{x^2-2x+1}=2-3\ln2$$
$$A=\int^0_{-1}\frac{x^3dx}{x^2-2x+1}$$ 1) Express the integrand as a sum of partial fractions: Since the power of $x$ in the numerator is higher than that in the denominator, we must divide the denominator into the numerator first, which gives us a polynomial and a proper fraction: $$\frac{x^3}{x^2-2x+1}=x+2+\frac{3x-2}{x^2-2x+1}$$ Now we concentrate on the fraction: $$\frac{3x-2}{x^2-2x+1}=\frac{3x-2}{(x-1)^2}=\frac{A}{x-1}+\frac{B}{(x-1)^2}$$ Clear fractions: $$A(x-1)+B=3x-2$$ $$Ax-A+B=3x-2$$ Equating coefficients of corresponding powers of $x$, we get - $A=3$ - $B-A=-2$ Solving for $B$, we get $B=1$. Therefore, $$\frac{3x-2}{x^2-2x+1}=\frac{3}{x-1}+\frac{1}{(x-1)^2}$$ 2) Evaluate the integral: $$A=\int^0_{-1}\Big[(x+2)+\frac{3}{x-1}+\frac{1}{(x-1)^2}\Big]dx$$ $$A=\frac{x^2}{2}+2x+3\ln|x-1|-\frac{1}{x-1}\Big]^0_{-1}$$ $$A=\Big(0+0+3\ln1+1\Big)-\Big(\frac{1}{2}-2+3\ln2+\frac{1}{2}\Big)$$ $$A=1-\Big(-1+3\ln2\Big)$$ $$A=2-3\ln2$$