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

$$\int^1_0\frac{x^3dx}{x^2+2x+1}=3\ln2-2$$
$$A=\int^1_0\frac{x^3dx}{x^2+2x+1}$$ 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 - $A=3$ - $A+B=2$ 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: $$A=\int^1_0\Big[(x-2)+\frac{3}{x+1}-\frac{1}{(x+1)^2}\Big]dx$$ $$A=\frac{x^2}{2}-2x+3\ln|x+1|+\frac{1}{x+1}\Big]^1_0$$ $$A=\Big(\frac{1}{2}-2+3\ln2+\frac{1}{2}\Big)-\Big(0-0+3\ln1+1\Big)$$ $$A=3\ln2-1-1=3\ln2-2$$