JEE Main 2015 · 11 Apr · Shift 1 · Mathematics Q74

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MathematicsSingle correct+4 / -1Question 74 of 90

If $\displaystyle\int\frac{\log\left(t+\sqrt{1+t^2}\right)}{\sqrt{1+t^2}}dt=\frac12(g(t))^2+C$ , where C is a constant, then g(2) is equal to :

2\log\left(2+\sqrt5\right)

\log\left(2+\sqrt5\right)

\dfrac1{\sqrt5}\log\left(2+\sqrt5\right)

\dfrac12\log\left(2+\sqrt5\right)

Answer and worked solution

Official answer: B

With $u=\log\left(t+\sqrt{1+t^2}\right)$ , $du=\dfrac{dt}{\sqrt{1+t^2}}$ , the integral is $\frac12u^2+C$ . So $g(t)=\log\left(t+\sqrt{1+t^2}\right)$ and $g(2)=\log\left(2+\sqrt5\right)$ .

Topics:Substitution

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JEE Main 2015 · 11 Apr · Shift 1, Mathematics question 74