Question

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