WBJEE · Maths · Indefinite Integration
If \(\int \frac{\log _e\left(x+\sqrt{\left.1+x^2\right)}\right.}{\sqrt{1+x^2}} d x=f(g(x))+c\) then
- A \(f(x)=\frac{x^2}{2}, g(x)=\log _e\left(x+\sqrt{1+x^2}\right)\)
- B \(f(x)=\log _e\left(x+\sqrt{1+x^2}\right), g(x)=\frac{x^2}{2}\)
- C \(f(x)=x^2, g(x)=\log _e\left(x+\sqrt{1+x^2}\right)\)
- D \(f(x)=\log _e\left(x-\sqrt{1+x^2}\right), g(x)=x^2\)
Answer & Solution
Correct Answer
(A) \(f(x)=\frac{x^2}{2}, g(x)=\log _e\left(x+\sqrt{1+x^2}\right)\)
Step-by-step Solution
Detailed explanation
Hint \(:\) Let \(\log \left(x+\sqrt{1+x^2}\right)=t \Rightarrow \frac{\frac{\sqrt{1+x^2}+x}{\sqrt{1+x^2}} d x=d t}{x+\sqrt{1+x^2}}\)…
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