Mathematics - Definite Integration Question with Solution | TestHub

To solve the integral, substitute $e^t-1=u^2$. This implies $e^{t}dt=2udu$, so $dt=\frac{2udu}{u^2+1}$.

The integral becomes:

$$I=\int \frac{2udu}{(u^2+1)u}=2\int \frac{du}{u^2+1}=2{\tan }^{-1}(u)$$

Substituting back $u=\sqrt{e^t-1}$:

$$I=2{\tan }^{-1}(\sqrt{e^t-1})$$

Now, evaluate the definite integral from $\log 2$ to $x$:

$$2{\tan }^{-1}(\sqrt{e^x-1})-2{\tan }^{-1}(\sqrt{e^{\log 2}-1})=\frac{\pi }{6}$$

$$2{\tan }^{-1}(\sqrt{e^x-1})-2{\tan }^{-1}(\sqrt{2-1})=\frac{\pi }{6}$$

$$2{\tan }^{-1}(\sqrt{e^x-1})-2{\tan }^{-1}(1)=\frac{\pi }{6}$$

$$2{\tan }^{-1}(\sqrt{e^x-1})-2\left(\frac{\pi }{4}\right)=\frac{\pi }{6}$$

$$2{\tan }^{-1}(\sqrt{e^x-1})-\frac{\pi }{2}=\frac{\pi }{6}$$

$$2{\tan }^{-1}(\sqrt{e^x-1})=\frac{\pi }{6}+\frac{\pi }{2}=\frac{\pi +3\pi }{6}=\frac{4\pi }{6}=\frac{2\pi }{3}$$

$${\tan }^{-1}(\sqrt{e^x-1})=\frac{\pi }{3}$$

$$\sqrt{e^x-1}=\tan \left(\frac{\pi }{3}\right)=\sqrt{3}$$

Squaring both sides:

$$e^x-1=3$$

$$e^x=4$$

Taking the natural logarithm of both sides:

$$x={\log }_{e}4$$

The final answer is $\boxed{{\log }_{e4}}$.