CUET · MATHS · PYQ PAPER 2023
The integral \(\int \frac{x^4+2}{x^2+1} d x\) is equal to
- A \(\frac{1}{3} x^3-x+3 \tan ^{-1} x+c\), where \(c\) is a constant
- B \(\frac{1}{3} x^3+x+3 \tan ^{-1} x+c\), where \(c\) is a constant
- C \(\frac{1}{3} x^3+x-3 \tan ^{-1} x+c\), where \(c\) is a constant
- D \(\frac{1}{3} x^3-x-3 \tan ^{-1} x+c\), where \(c\) is a constant
Answer & Solution
Correct Answer
(A) \(\frac{1}{3} x^3-x+3 \tan ^{-1} x+c\), where \(c\) is a constant
Step-by-step Solution
Detailed explanation
\(\int \frac{x^4+2}{x^2+1} d x = \int \frac{x^4-1+3}{x^2+1} d x\) \(= \int \frac{(x^2-1)(x^2+1)+3}{x^2+1} d x\) \(= \int \left( x^2-1 + \frac{3}{x^2+1} \right) d x\) \(\frac{1}{3} x^3-x+3 \tan ^{-1} x+c\)
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