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JEE Mains · Maths · STD 12 - 7.1 indefinite integral

यदि \(\int \frac{\log \left(t+\sqrt{1+t^{2}}\right)}{\sqrt{1+t^{2}}} d t=\frac{1}{2}(g(t))^{2}+C\) है, जहाँ \(C\) एक अचर है, तो \(g (2)\) बराबर है

  1. A \(\frac{1}{{\sqrt 5 }}\log \left( {2 + \sqrt 5 } \right)\)
  2. B \(\frac{1}{2}\log \left( {2 + \sqrt 5 } \right)\)
  3. C \(2\log \left( {2 + \sqrt 5 } \right)\)
  4. D \(\log \left( {2 + \sqrt 5 } \right)\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\log \left( {2 + \sqrt 5 } \right)\)

Step-by-step Solution

Detailed explanation

Let \(I=\int \frac{\log (t+\sqrt{1+t^{2}})}{\sqrt{1+t^{2}}} d t\) put \(u=\log (t+\sqrt{1+t^{2}})\) \(\mathrm{du}=\frac{1}{\mathrm{t}+\sqrt{1+\mathrm{t}^{2}}} \cdot\left[\frac{\mathrm{t}+\sqrt{1+\mathrm{t}^{2}}}{\sqrt{1+\mathrm{t}^{2}}}\right]\) \(=\frac{1}{\sqrt{1+t^{2}}} d t\)…
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