# 8.5-Method of Partial Fractions-Advanced ## Anna Smith ### 11/12/2020 --- Question 4: By means of the partial fraction decomposition method, compute the integral: $\int\frac{x^3+x+1}{(x^2+1)(x^2+x+1)}dx$ --- $\int\frac{x^3+x+1}{(x^2+1)(x^2+x+1)}dx=\frac{Ax+B}{(x^2+1)}+\frac{Cx+D}{x^2+x+1}$ $x^3+x+1=(Ax+B)(x^2+x+1)+(Cx+D)(x^2+1)$ $x^3+x+1=$ $x^3A+x^3C+x^2A+x^2B+x^2D+xA+xB+xC+B+D$ Add like terms: --- $x^3+x+1=$ $x^3(A+C)+x^2(A+B+D)+x(A+B+C)+B+D$ Set $x=0$ $1=B+D\rightarrow D=1-B$ $D=1$ Set $x=1$ $3=3(A+B)+2(C+D)$ $3=3A+3B+2C+2-2B\rightarrow 1=3A+B+2C$ $C=2$ --- Set $x=-1$ $-1=B-A+2(0-C)$ $-1=B-A+2-2B-1+3A+B$ $-2=2A \rightarrow A=-1$ Set $x=2$ $11=7(2A+B)+5(2C+D)\rightarrow 11=-14+7B+5+15$ $\rightarrow -5B+5-5B\rightarrow 0=-3B$ $\rightarrow B=0$ --- $\int\frac{x^3+x+1}{(x^2+1)(x^2+x+1)}dx=\frac{-1x+0}{(x^2+1)}+\frac{2x+1}{(x^2+x+1)}$ $=\frac{-x}{(x^2+1)}+\frac{2x+1}{(x^2+x+1)}$ $=\int\frac{-x}{(x^2+1)}+\frac{2x+1}{(x^2+x+1)}$ --- Take the antiderivative: $u=x^2+1,du=2xdx,\frac{-1}{2}du=-xdx$ $\frac{-1}{2}\int\frac{1}{u}\rightarrow \frac{-1}{2}ln(u)=\frac{-1}{2}ln(x^2+1)$ $u=x^2+x+1,du=2x+1dx$ $\rightarrow \int\frac{1}{u} \rightarrow ln(u)=ln(x^2+x+1)$ Final answer... $\frac{-1}{2}ln(x^2+1)+ln(x^2+x+1)$