If (t+√1+t^2)√1+t^2,dt = 12(g(t))^2+C, where C is a constant, then g(b) is equal to: (Here…

Question

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

(Here $b=2$, per the worked solution.)

Accepted Answer

Let $I = \int \dfrac{\log(t+\sqrt{1+t^2})}{\sqrt{1+t^2}}\,dt$

Put $u = \log(t+\sqrt{1+t^2})$, then $du = \dfrac{1}{\sqrt{1+t^2}}\,dt$

$I = \int u\,du = \dfrac{u^2}{2}+c$

Since $I = \dfrac12[g(t)]^2+c$, $g(t) = \log(t+\sqrt{1+t^2})$

Put $t=2$: $g(2) = \log(2+\sqrt5)$

Solutions · 1