91Ó°ÊÓ

Replace \(1+2x\) with \(u^2\), and express \(dx\) in terms of \(du\). To find the expression for \(dx\), differentiate the equation \(1+2x = u^2\) with respect to \(x\) to get: $$2\,dx = 2u\,du$$ Now, divide both sides by 2: $$dx = u\,du$$ Now, substitute these expressions in the integral: $$\int \frac{d x}{x \sqrt{1+2 x}} = \int \frac{u\,du}{x\,u} $$

Notice that the \(u\) in the numerator and denominator of the integrand cancel each other out: $$\int \frac{u\,du}{x\,u}= \int \frac{du}{x}$$ Now, express x in terms of u using the substitution equation: $$x = \frac{u^2 - 1}{2}$$ Substitute this into the integral: $$\int \frac{du}{x} = \int \frac{du}{\frac{u^2 - 1}{2}}$$

To proceed with evaluating the integral, multiply the numerator and denominator by 2: $$\int \frac{du}{\frac{u^2 - 1}{2}} = \int \frac{2\,du}{u^2-1}$$ This integral can now be evaluated using the following trigonometrical substitution: Take \(u=\cosh{t}\), where \(t\) is the hyperbolic angle. Then, \(du = \sinh{t}\:dt\). The integral becomes: $$\int \frac{2\,dt}{\cosh^2{t}-1}$$ Notice that \(\cosh^2{t}- \sinh^2{t} = 1\), so we have: $$\int \frac{2\,dt}{\sinh{t}\:\sinh{t}}$$ This simplifies to: $$\int \frac{2\,dt}{\sinh^2{t}} = 2\int \cosh^2{t}\,dt$$ Now, use the formula: $$\cosh^2{t} = \frac{1+\cosh{2t}}{2}$$ So, the integral becomes: $$2\int\frac{1+\cosh{2t}}{2}\,dt$$ Now, distribute the integral: $$\int 1\,dt + \int\cosh{2t}\,dt$$ Integrate each term: $$t + \frac{1}{2}\sinh{2t} + C$$ Finally, bring back the original variable \(x\). Recall that \(u=\cosh{t}\), so \(t=\cosh^{-1}{u}\). And from the original substitution: \(1+2x=u^2\). Thus, the final answer is: $$\cosh^{-1}(\sqrt{1+2x}) + \frac{1}{2}\sinh(2\cosh^{-1}(\sqrt{1 + 2x})) + C$$