evaluate the integral. ∫ 13 / ((x - 3)(x² + 4)) dx

Explanation:

Step1: Decompose into partial - fractions

Let $\frac{13}{(x - 3)(x^{2}+4)}=\frac{A}{x - 3}+\frac{Bx + C}{x^{2}+4}$. Then $13=A(x^{2}+4)+(Bx + C)(x - 3)$. Set $x = 3$, we get $13=A(9 + 4)$, so $A = 1$. Expand the right - hand side: $13=Ax^{2}+4A + Bx^{2}-3Bx + Cx-3C=(A + B)x^{2}+(-3B + C)x+(4A-3C)$. Since $A = 1$, then $A + B=0$ gives $B=-1$, and $4A-3C = 13$ (substitute $A = 1$) gives $4-3C = 13$, so $C=-3$. So $\frac{13}{(x - 3)(x^{2}+4)}=\frac{1}{x - 3}+\frac{-x - 3}{x^{2}+4}=\frac{1}{x - 3}-\frac{x}{x^{2}+4}-\frac{3}{x^{2}+4}$.

Step2: Integrate term - by - term

$\int\frac{13}{(x - 3)(x^{2}+4)}dx=\int\frac{1}{x - 3}dx-\int\frac{x}{x^{2}+4}dx-\int\frac{3}{x^{2}+4}dx$. For $\int\frac{1}{x - 3}dx=\ln|x - 3|+C_1$. For $\int\frac{x}{x^{2}+4}dx$, let $u=x^{2}+4$, $du = 2xdx$, then $\int\frac{x}{x^{2}+4}dx=\frac{1}{2}\int\frac{du}{u}=\frac{1}{2}\ln|u|=\frac{1}{2}\ln(x^{2}+4)+C_2$. For $\int\frac{3}{x^{2}+4}dx$, since $\int\frac{1}{a^{2}+x^{2}}dx=\frac{1}{a}\arctan(\frac{x}{a})+C$ ($a = 2$ here), then $\int\frac{3}{x^{2}+4}dx=\frac{3}{2}\arctan(\frac{x}{2})+C_3$.

Step3: Combine the results

$\int\frac{13}{(x - 3)(x^{2}+4)}dx=\ln|x - 3|-\frac{1}{2}\ln(x^{2}+4)-\frac{3}{2}\arctan(\frac{x}{2})+C$, where $C=C_1 - C_2 - C_3$.

Answer:

$\ln|x - 3|-\frac{1}{2}\ln(x^{2}+4)-\frac{3}{2}\arctan(\frac{x}{2})+C$