ExamBro
ExamBro
MHT CET · Maths · Indefinite Integration

\(\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\frac{1}{2}(g(x))^2+C\), (where \(C\) is constant of integration.) Then \(g(x)=\)

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

Answer & Solution

Correct Answer

(A) \(\log \left(x+\sqrt{1+x^2}\right)\)

Step-by-step Solution

Detailed explanation

\(\int \frac{\log \left(x+\sqrt{1+x^2}\right)}{\sqrt{1+x^2}} \mathrm{~d} x=\int t \mathrm{~d} t=\frac{t^2}{2}+C\)
[where \(t=\log \left(x+\sqrt{1+x^2}\right)\) ]
\(\begin{aligned}
& =\frac{\left\{\log \left(x+\sqrt{1+x^2}\right)\right\}^2}{2}+C \\
& \Rightarrow g(x)=\log \left(x+\sqrt{1+x^2}\right)^2
\end{aligned}\)