JEE Mains · Maths · STD 12 - 7.1 indefinite integral
If \(\int {\frac{{log\left( {t + \sqrt {1 + {t^2}} } \right)}}{{\sqrt {1 + {t^2}} }}dt = \frac{1}{2}{{\left( {g\left( t \right)} \right)}^2} + C} \) , where \(C\) is a constant, then \(g(2)\) is equal to
- A \(\frac{1}{{\sqrt 5 }}\log \left( {2 + \sqrt 5 } \right)\)
- B \(\frac{1}{2}\log \left( {2 + \sqrt 5 } \right)\)
- C \(2\log \left( {2 + \sqrt 5 } \right)\)
- D \(\log \left( {2 + \sqrt 5 } \right)\)
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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