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GUJCET · Maths · Integrals
\(\int_1^3\left(\frac{x^2+1}{4 x}\right)^{-1} d x=\) __________.
- A \(\log 5\)
- B \(\frac{1}{2} \log 5\)
- C \(\log 25\)
- D \(\log 100\)
Answer & Solution
Correct Answer
(C) \(\log 25\)
Step-by-step Solution
Detailed explanation
\(\int_1^3\left(\frac{x^2+1}{4 x}\right)^{-1} d x = \int_1^3 \frac{4x}{x^2+1} d x\) \(= [2 \log(x^2+1)]_1^3\)
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