JEE Mains · Maths · STD 12 - 7.2 definite integral
The integral \(\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right) d x}{\left(2+x^{2}\right) \sqrt{4+x^{4}}}\) is equal to
- A \(1\)
- B \(2\)
- C \(4\)
- D \(3\)
Answer & Solution
Correct Answer
(D) \(3\)
Step-by-step Solution
Detailed explanation
\(\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{\left(2-x^{2}\right)}{\left(x^{2}+2\right) \sqrt{4+x^{4}}} d x\) \(\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{x^{2}\left(\frac{2}{x^{2}}-1\right) d x}{x\left(x+\frac{2}{x}\right) \times x \sqrt{\frac{4}{x^{2}}+x^{2}}}\)…
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