Question

If \(\displaystyle\int\dfrac{\log\left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}}\,dt=\dfrac12\left(g(t)\right)^2+C\), where C is a constant, then \(g(2)\) is equal to :

• A. \(2\log\left(2+\sqrt5\right)\)
• B. \(\log\left(2+\sqrt5\right)\)
• C. \(\dfrac{1}{\sqrt5}\log\left(2+\sqrt5\right)\)
• D. \(\dfrac12\log\left(2+\sqrt5\right)\)

Hint:

Substitute \(u=\log\left(t+\sqrt{1+t^2}\right)\); then \(du=\frac{dt}{\sqrt{1+t^2}}\).

Answer 1

The Correct Option is B

Let \(u=\log\left(t+\sqrt{1+t^2}\right)\). Then \(du=\dfrac{1+\frac{t}{\sqrt{1+t^2}}}{t+\sqrt{1+t^2}}dt=\dfrac{dt}{\sqrt{1+t^2}}\).
The integral becomes \(\int u\,du=\frac{u^2}{2}+C\), so \(g(t)=\log\left(t+\sqrt{1+t^2}\right)\) (up to sign).
\(g(2)=\log\left(2+\sqrt5\right)\).
Answer: (2) \(\log\left(2+\sqrt5\right)\)