COMEDK · Maths · 28. Indefinite Integration
\(\int \dfrac{dx}{x\sqrt{x^2 + 4}} =\)
- A \(\dfrac{1}{4}\log\left|\dfrac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}-2}\right| + C\)
- B \(\dfrac{1}{2}\log\left|\dfrac{\sqrt{x^2+4}+2}{\sqrt{x^2+4}-2}\right| + C\)
- C \(\dfrac{1}{4}\log\left|\dfrac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right| + C\)
- D \(\dfrac{1}{2}\log\left|\dfrac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right| + C\)
Answer & Solution
Correct Answer
(C) \(\dfrac{1}{4}\log\left|\dfrac{\sqrt{x^2+4}-2}{\sqrt{x^2+4}+2}\right| + C\)
Step-by-step Solution
Detailed explanation
Let \(I = \int \dfrac{dx}{x\sqrt{x^2+4}}\) Multiplying the numerator and the denominator by \(x\), we get: \(I = \int \dfrac{x}{x^2\sqrt{x^2+4}} dx\) Let \(x^2+4 = t^2\). Differentiating both sides, we get: \(2x dx = 2t dt \Rightarrow x dx = t dt\) Also, \(x^2 = t^2-4\).…
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