TS EAMCET · Maths · Indefinite Integration
\(\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=\)
- 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 & Solution
Correct Answer
(C) \(\frac{x^5}{5}+x^3+6 x+c\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { } \int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x \\ & \because \frac{x^8-9 x^2+18}{x^4-3 x^2+3}=x^4+3 x^2+\frac{6 x^4-18 x^2+18}{x^4-3 x^2+3} \\ & \Rightarrow \quad x^4+3 x^2+6 \end{aligned} \] Now,…
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