91Ó°ÊÓ

First, let us evaluate \(\int \frac{\sqrt{2 x+1}}{x^{2}} d x\). Let \(u = 2x + 1\), which means that \(du = 2 dx\). Now let's rewrite the integral in terms of \(u\): \(\int \frac{\sqrt{2 x+1}}{x^{2}} d x = \int \frac{\sqrt{u}}{((u-1)/2)^{2}}\cdot\frac{1}{2}du\) Now, simplifying the expression gives: \(= \frac{1}{2}\int \frac{2u\sqrt{u}}{(u-1)^{2}} du\) Integrating with respect to \(u\), we get: \(= \frac{1}{2}\int 2u^{3/2}\left(1 -\frac{1}{u}\right)^{2} du\) Now, we can use integration by substitution again with \(v = 1 -\frac{1}{u}\), \(dv = \frac{1}{u^{2}} du\). Thus, \(= \int u^{3/2} vd(v)= \int (1-v)^{3/2} vd(v)\) Integrating this equation is not possible with elementary functions. Therefore, the integral does not have an elementary antiderivative. You can express the result in terms of special functions, but for this exercise, we won't go into such details.

Next, we will evaluate \(\int \frac{x d x}{(a+b x)^{1 / 2}}\). Let \(u = a + bx\), so \(du = b dx\). Dividing by \(b\), we get \(\frac{1}{b} du = dx\). Substituting the variable \(u\) into the integral, we get: \(\int \frac{x d x}{(a+b x)^{1 / 2}}=\int \frac{\frac{u-a}{b} }{u^{1 / 2}}\cdot\frac{1}{b} du\) Simplifying the expression, we obtain: \(= \frac{1}{b^{2}}\int \frac{u-a}{\sqrt{u}} du\) Now, we can split the expression into two simpler integrals: \(= \frac{1}{b^{2}}\left(\int u\sqrt{u}^{-1} du - a\int u^{-1/2} du \right)\) Integrating both of them will give: \(= \frac{1}{b^{2}}\left(2u^{\frac{1}{2}} - 2a\sqrt{u} \right) + C\) Replacing \(u\) by \(a + bx\), we have: \(= \frac{1}{b^{2}}\left(2(a+bx)^{\frac{1}{2}}-2a\sqrt{a+bx}\right) + C\)

Finally, let's evaluate \(\int \sqrt{\frac{x+a}{x+b}} d x\). For this integral, perform the substitution \(u^2=\frac{x+a}{x+b}\). Then, \((x+a)=u^2(x+b)\) and \(x=(\frac{u^2-a}{u^2-1})b\). We can compute \(dx\) as \(\frac{2ab(1-u^2)}{(1-u^2)^2}du\). Replacing \(x\) and \(dx\) in the integrand: \(\int \sqrt{\frac{x+a}{x+b}} d x = \int \frac{2ab(1-u^2)}{(1-u^2)^2}u du\) Simplifying, we are left with: \(= 2ab\int \frac{u (1-u^2)}{(1-u^2)^{2}} du\) Use the substitution \(v = 1 - u^2\), and \(dv = -2u du\). Thus, the integral can be rewritten as: \(= 2ab\int \frac{-\frac{1}{2}dv}{v^{2}}\) Integrating this expression, we have: \(= ab\left(\frac{1}{v}\right) + C\) Replacing \(v\) by \(1-u^2\), and \(u^2\) by \(\frac{x+a}{x+b}\), we have: \(= ab\left(\frac{1}{1-\left(\frac{x+a}{x+b}\right)}\right) + C\) Simplifying, we obtain the final result for this integral: \(= ab\left(\frac{x+b}{x}\right) + C\)