AP EAMCET · Maths · Indefinite Integration
\(\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x=\)
- A \(\frac{x^5}{4}+x^3+6 x^2+c\)
- 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
\(\int \frac{x^8-9 x^2+18}{x^4-3 x^2+3} d x\) degree of numerator is greater than degree of denominator so, \(x^8-9 x^2+18\) is divided by \(x^4-3 x^2+3\) We get quotient \(x^4+3 x^2+6\) and 0 remainder, so…
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