Question

If the primitive of $\frac{x^5+x^4-8}{x^3-4x}$ is $\frac{x^3}{3}+\frac{x^2}{2}+Ax+\left|\log f\left(x\right)\right|+C$ then

• A. $A=1$
• B. $A=4$
• C. $f\left(x\right)=x^2(x-2{)}^5(x+2{)}^{-3}$
• D. $f\left(x\right)=x^2(x-2{)}^3(x+2{)}^{-2}$

Answer 1

Answer: (B), (C)

$A=4$
$f\left(x\right)=x^2(x-2{)}^5(x+2{)}^{-3}$

$\frac{x^5+x^4-8}{x^3-4x}=x^2+x+4+\frac{4(x^2+4x-2)}{x(x^2-4)}$
$=x^2+x+4+\frac{4(x^2-4+4x+2)}{x(x^2-4)}$
$=x^2+x+4+\frac{4}{x}+\frac{16}{(x^2-4)}+\frac{8}{x(x^2-4)}$
$=x^2+x+4+\frac{2}{x}+\frac{5}{x-2}-\frac{3}{x-2}$
Hence the primitive of the given function is
$\frac{x^3}{3}+\frac{x^2}{2}+4x+2\log \left|x\right|+5\log |x-2|-3\log |x-2|+c$
$=\frac{x^3}{3}+\frac{x^2}{2}+4x+\log \left|x^2{(x-2)}^5{(x+2)}^{-3}\right|+c$