Question

if $\int \frac{3x+4}{x^3-2x-4}dx=log\left|x-2\right|+klogf\left(x\right)+c$, then

• A. $f\left(x\right)=f\left(x\right)=\left|x^2+2x+2\right|$
• B. $f\left(x\right)=x^2+2x+2$
• C. $k=-1/2$
• D. All of these

Answer 1

Answer: (D)

$\frac{3x+4}{x^3-2x-4}=\frac{3x+4}{\left(x-2\right)\left(x^2+2x+2\right)}$

$=\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+2}$

$\Rightarrow 3x+4=A\left(x^2+2x+2\right)+\left(Bx+C\right)\left(x-2\right)$

$\therefore A+B=O^{\prime },2A-2B+C=3,2A-2C=4,\Rightarrow A=1,B=C=-1$

$\therefore \int \frac{3x+4}{x^3-2x-4}dx=\int \frac{dx}{x-2}-\frac{1}{2}\int \frac{2x+2}{x^2+2x+2}dx$

$=log\left|x-2\right|-\frac{1}{2}log\left|x^2+2x+2\right|+C$

$\Rightarrow k=-\frac{1}{2}andf\left(x\right)=\left|x^2+2x+2\right|$