19. ∫(7x/((2x - 3)(x + 2)))dx = (a) (3/2)ln|2x - 3| + 2ln|x + 2| + c (b) 3ln|2x - 3| + 2ln|x + 2| + c (c)…

Explanation:

Step1: Decompose into partial - fractions

Let $\frac{7x}{(2x - 3)(x + 2)}=\frac{A}{2x - 3}+\frac{B}{x + 2}$. Then $7x=A(x + 2)+B(2x - 3)$. Set $x=-2$, we get $7\times(-2)=A(-2 + 2)+B(2\times(-2)-3)$, so $-14=B(-4 - 3)$, and $B = 2$. Set $x=\frac{3}{2}$, we get $7\times\frac{3}{2}=A(\frac{3}{2}+2)+B(2\times\frac{3}{2}-3)$, so $\frac{21}{2}=A(\frac{3 + 4}{2})$, and $A = 3$. So $\frac{7x}{(2x - 3)(x + 2)}=\frac{3}{2x - 3}+\frac{2}{x + 2}$.

Step2: Integrate term - by - term

$\int\frac{7x}{(2x - 3)(x + 2)}dx=\int\frac{3}{2x - 3}dx+\int\frac{2}{x + 2}dx$. For $\int\frac{3}{2x - 3}dx$, let $u = 2x-3$, $du=2dx$, then $\int\frac{3}{2x - 3}dx=\frac{3}{2}\int\frac{du}{u}=\frac{3}{2}\ln|u|=\frac{3}{2}\ln|2x - 3|$. For $\int\frac{2}{x + 2}dx=2\ln|x + 2|$. So $\int\frac{7x}{(2x - 3)(x + 2)}dx=\frac{3}{2}\ln|2x - 3|+2\ln|x + 2|+C$.

Answer:

A. $\frac{3}{2}\ln|2x - 3|+2\ln|x + 2|+C$