Question

If $\int \frac{3x+4}{x^3-2x-4}dx=\log |x-2|+K\log f\left(x\right)+C$, then

• A. $K=-1/2$
• B. $f\left(x\right)=x^2+2x+2$
• C. $f\left(x\right)=|x^2+2x+2|$
• D. $K=1/4$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(x\right)=x^2+2x+2$
B $K=-1/2$
C $f\left(x\right)=|x^2+2x+2|$

Since $x^3-2x-4=(x-2)(x^2+2x+2)$

$\frac{3x+4}{x^3-2x-4}=\frac{A}{x-2}+\frac{Bx+C}{(x^2+2x+2)}$

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

Putting $x=2,A=1$ and comparing coefficients of $x^2$, we get

$0=A+B\Rightarrow B=-1$

Putting $x=0$ we have $4=2A-2C\Rightarrow C=-1$

Thus

$\int \frac{3x+4}{x^3-2x-4}dx=\int \frac{1dx}{x-2}-\int \frac{x+1}{(x^2+2x+2)}dx$
$=\log |x-1|-\frac{1}{2}\log \left|(x^2+2x+2)\right|+c$

Comparing with , $\int \frac{3x+4}{x^3-2x-4}dx=\log |x-2|+K\log f\left(x\right)+C$
Hence $K=-\frac{1}{2}$

and $f\left(x\right)=\left|(x^2+2x+2)\right|=(x^2+2x+2)$ as $(x^2+2x+2)>0$