Question

$\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x=$

• A. $\frac{x^{4}}{4}+x^{3}+6 x^{2}+c$
• B. $\frac{x^{5}}{5}+\frac{x^{4}}{4}+6 x+c$
• C. $\frac{x^{5}}{5}+x^{3}+6 x+c$
• D. $\frac{x^{5}}{5}-\frac{x^{3}}{2}+6 x^{2}+c$

Answer 1

Answer: (C)

$\int \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3} d x$
$\because\, \frac{x^{8}-9 x^{2}+18}{x^{4}-3 x^{2}+3}-x^{4}+3 x^{2}+\frac{6 x^{4}-18 x^{2}+18}{x^{4}-3 x^{2}+3}$
$\Rightarrow \, x^{4}+3 x^{2}+6$
Now, $\int x^{4}+3 x^{2}+6 d x$
$ \Rightarrow \, \frac{x^{5}}{5}+\frac{3 x^{3}}{3}+6 x+c$
$\Rightarrow \, \frac{x^{5}}{5}+x^{3}+6 x+c$