Question

If $\int \frac{9x+15}{x^3-6x-9}dx=A\log |g(x)|+B\log \mid f(x)+C$, then $\frac{(A-B)g(4)}{f(-1)}=$

• A. $3$
• B. $\frac{1}{7}$
• C. $1$
• D. $\frac{3}{7}$

Answer 1

Answer: (A)

We have, $\int \frac{9x+15}{x^3-6x-9}dx=A\log |g(x)|+B\log |f(x)|+C$
Now, $\frac{9x+15}{x^3-6x-9}=\frac{9x+15}{(x-3)\left(x^2+3x+3\right)}$
$\therefore \frac{9x+15}{(x-3)\left(x^2+3x+3\right)}=\frac{A}{x-3}+\frac{Bx+C}{x^2+3x+3}$
$\Rightarrow 9x+15=A\left(x^2+3x+3\right)+(Bx+c)(x-3)$
$\Rightarrow 9x+15=x^2(A+B)+x(3A-3B+C)+(3A-3C)$
On equating coefficients of $x^2,x$ and constant term, we get
$A+B=0,3A-3B+C=9$ and $3A-3C=15$
On solving above three equations
We get, $A=2,B=-2$ and $C=-3$
$\therefore \int \frac{9x+15}{(x-3)\left(x^2+3x+3\right)}dx$
$=\int \frac{2}{x-3}dx+\int \frac{-2x-3}{x^2+3x+3}dx$
$=\int \frac{2}{x-3}dx-\int \frac{2x+3}{x^2+3x+3}dx$
$=2\log |x-3|-\log \left|x^2+3x+3\right|+C$
$\therefore A=2,B=-1,g(x)=x-3$ and $f(x)=x^2+3x+3$
$\therefore \frac{(A-B)g(4)}{f(-1)}=\frac{(2+1)(4-3)}{(-1{)}^2+3(-1)+3}=3$