Q113 / 160 · TS EAMCET 2018 (04 May Shift 2) · Mathematics

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

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

Solution

$$ \begin{aligned} & \text { } \int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x \\ & \because \frac{x^8-9 x^2+18}{x^4-3 x^2+3}=x^4+3 x^2+\frac{6 x^4-18 x^2+18}{x^4-3 x^2+3} \\ & \Rightarrow \quad x^4+3 x^2+6 \end{aligned} $$ Now, $\int x^4+3 x^2+6 d x \Rightarrow \frac{x^5}{5}+\frac{3 x^3}{3}+6 x+c$ $$ \Rightarrow \quad \frac{x^5}{5}+x^3+6 x+c $$

Hence, the correct answer is option: 'C'.

Relevant Questions