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 \(g(x)=\)
- A \(\log \left(x+\sqrt{1+x^2}\right)\)
- B \(\log \left(x+\sqrt{1+2 x^2}\right)\)
- C \(\log \left(x-\sqrt{1+x^2}\right)\)
- D \(\log \left(\sqrt{1+x^2}\right)\)
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 \(t=\log \left(x+\sqrt{1+x^2}\right)\) ]
\(\begin{aligned}
& =\frac{\left\{\log \left(x+\sqrt{1+x^2}\right)\right\}^2}{2}+C \\
& \Rightarrow g(x)=\log \left(x+\sqrt{1+x^2}\right)^2
\end{aligned}\)
[where \(t=\log \left(x+\sqrt{1+x^2}\right)\) ]
\(\begin{aligned}
& =\frac{\left\{\log \left(x+\sqrt{1+x^2}\right)\right\}^2}{2}+C \\
& \Rightarrow g(x)=\log \left(x+\sqrt{1+x^2}\right)^2
\end{aligned}\)
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