Question

evaluate the integral.\\(\int \frac{\ln \sqrt{8x}}{x} dx\\)\\(\int \frac{\ln \sqrt{8x}}{x} dx = (\ln \sqrt{8x})^2 + c\\)(use parentheses to clearly denote the argument of each function.)

Explanation:

Step1: Simplify the Integrand

First, simplify \(\ln\sqrt{8x}\). Recall that \(\ln\sqrt{a}=\frac{1}{2}\ln a\), so \(\ln\sqrt{8x}=\frac{1}{2}\ln(8x)=\frac{1}{2}(\ln 8+\ln x)\). But maybe a substitution is better. Let \(u = \ln\sqrt{8x}\), then find \(du\).

Step2: Substitution

Let \(u=\ln\sqrt{8x}\). Then, differentiate \(u\) with respect to \(x\). Using the chain rule: \(\frac{du}{dx}=\frac{1}{\sqrt{8x}}\cdot\frac{1}{2\sqrt{8x}}\cdot 8\)? Wait, no, simpler: \(\sqrt{8x}=(8x)^{\frac{1}{2}}\), so \(\ln\sqrt{8x}=\frac{1}{2}\ln(8x)\). Then \(\frac{d}{dx}(\ln\sqrt{8x})=\frac{1}{2}\cdot\frac{8}{8x}=\frac{1}{2x}\)? Wait, no, \(\frac{d}{dx}(\ln(8x))=\frac{8}{8x}=\frac{1}{x}\), so \(\frac{d}{dx}(\frac{1}{2}\ln(8x))=\frac{1}{2x}\). Wait, but the integrand is \(\frac{\ln\sqrt{8x}}{x}dx\). Let's do substitution properly. Let \(u = \ln\sqrt{8x}\), then \(du=\frac{1}{\sqrt{8x}}\cdot\frac{1}{2\sqrt{8x}}\cdot 8 dx\)? No, better: \(u = \ln\sqrt{8x}\), so \(u=\frac{1}{2}\ln(8x)\). Then \(du=\frac{1}{2}\cdot\frac{8}{8x}dx=\frac{1}{2x}dx\)? Wait, no, \(\frac{d}{dx}(\ln(8x))=\frac{8}{8x}=\frac{1}{x}\), so \(\frac{d}{dx}(\frac{1}{2}\ln(8x))=\frac{1}{2x}\). But the integrand is \(\frac{\ln\sqrt{8x}}{x}dx=\frac{u}{x}dx\). Wait, from \(u = \ln\sqrt{8x}\), we can also note that \(\frac{du}{dx}=\frac{1}{\sqrt{8x}}\cdot\frac{1}{2\sqrt{8x}}\cdot 8\)? No, that's wrong. Let's compute \(du\) correctly. Let \(y = \sqrt{8x}=(8x)^{\frac{1}{2}}\), then \(\ln y\) has derivative \(\frac{1}{y}\cdot y' \). \(y'=\frac{1}{2}(8x)^{-\frac{1}{2}}\cdot 8=\frac{4}{\sqrt{8x}}=\frac{4}{(8x)^{\frac{1}{2}}}=\frac{4}{2\sqrt{2x}}=\frac{2}{\sqrt{2x}}=\sqrt{\frac{2}{x}}\). Wait, this is getting complicated. Alternatively, notice that the integrand is \(\frac{\ln\sqrt{8x}}{x}dx\). Let \(u=\ln\sqrt{8x}\), then \(du=\frac{1}{\sqrt{8x}}\cdot\frac{1}{2\sqrt{8x}}\cdot 8 dx\)? No, let's use the power rule for logarithms: \(\ln\sqrt{8x}=\frac{1}{2}\ln(8x)\), so the integrand is \(\frac{\frac{1}{2}\ln(8x)}{x}dx=\frac{1}{2}\cdot\frac{\ln(8x)}{x}dx\). Now, let \(v = \ln(8x)\), then \(dv=\frac{8}{8x}dx=\frac{1}{x}dx\). So the integral becomes \(\frac{1}{2}\int v dv\), which is \(\frac{1}{2}\cdot\frac{v^2}{2}+C=\frac{v^2}{4}+C\). But \(v = \ln(8x)=2\ln\sqrt{8x}\), so \(v^2 = 4(\ln\sqrt{8x})^2\), so \(\frac{v^2}{4}=(\ln\sqrt{8x})^2\), which matches the given answer. Wait, actually, if we let \(u=\ln\sqrt{8x}\), then \(du=\frac{1}{x}dx\)? Wait, no, let's check: \(u=\ln\sqrt{8x}\), then \(u=\frac{1}{2}\ln(8x)\), so \(du=\frac{1}{2}\cdot\frac{8}{8x}dx=\frac{1}{2x}dx\)? No, that's not. Wait, maybe the original answer is using a substitution where \(u=\ln\sqrt{8x}\), and then \(du=\frac{1}{x}dx\)? Wait, no, let's compute \(\frac{d}{dx}(\ln\sqrt{8x})\):

\(\ln\sqrt{8x}=\frac{1}{2}\ln(8x)\), so derivative is \(\frac{1}{2}\cdot\frac{8}{8x}=\frac{1}{2x}\). But the integrand is \(\frac{\ln\sqrt{8x}}{x}dx=\frac{u}{x}dx\). If we let \(u=\ln\sqrt{8x}\), then \(x du=\frac{1}{2}dx\)? No, this is confusing. Wait, the given answer is \((\ln\sqrt{8x})^2 + C\). Let's differentiate the answer to check. Let \(F(x)=(\ln\sqrt{8x})^2 + C\). Then \(F'(x)=2\ln\sqrt{8x}\cdot\frac{1}{\sqrt{8x}}\cdot\frac{1}{2\sqrt{8x}}\cdot 8\)? No, simpler: \(F(x)=(\frac{1}{2}\ln(8x))^2 + C=\frac{1}{4}(\ln(8x))^2 + C\). Then \(F'(x)=\frac{1}{4}\cdot 2\ln(8x)\cdot\frac{8}{8x}=\frac{1}{4}\cdot 2\ln(8x)\cdot\frac{1}{x}=\frac{\ln(8x)}{2x}\). But the integrand is \(\frac{\ln\sqrt{8x}}{x}=\frac{\frac{1}{2}\ln(8x)}{x}=\frac{\ln(8x)}{2x}\), which matches \(F'(x)\). So the differentiation of the answer gives the integrand, so the answer is correct.

Step3: Verify by Differentiation

Differentiate \((\ln\sqrt{8x})^2 + C\) with respect to \(x\). Let \(u = \ln\sqrt{8x}\), so the derivative is \(2u\cdot u'\). \(u'=\frac{1}{\sqrt{8x}}\cdot\frac{1}{2\sqrt{8x}}\cdot 8\)? No, as before, \(u=\frac{1}{2}\ln(8x)\), so \(u'=\frac{1}{2}\cdot\frac{8}{8x}=\frac{1}{2x}\). Then \(2u\cdot u' = 2\cdot\frac{1}{2}\ln(8x)\cdot\frac{1}{2x}=\frac{\ln(8x)}{2x}\), but the integrand is \(\frac{\ln\sqrt{8x}}{x}=\frac{\frac{1}{2}\ln(8x)}{x}=\frac{\ln(8x)}{2x}\), which matches. So the differentiation confirms that the antiderivative is correct.

Answer:

\(\boxed{(\ln\sqrt{8x})^2 + C}\)