Evaluate { {x}}{3 x} \, dx

Correct Answer is : 13[log⁡(x)]2+c1/3 [ ({x})]² + c31[log(x)]2+c

Correct Answer is : 13[log⁡(x)]2+c1/3 [ ({x})]² + c31[log(x)]2+c

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Integral Calculus

Question : 32 of 73

Marks: +1, -0

Evaluate

\int \frac{\log \sqrt{x}}{3 x} \, dx

Solution:

Concept:

Simplify the integrand using properties of logarithms and then use substitution.

Explanation:

Let

I = \int \frac{\log \sqrt{x}}{3x} \, dx

Since

\log \sqrt{x} = \frac{1}{2} \log x

, we have

I = \int \frac{\frac{1}{2} \log x}{3x} \, dx = \frac{1}{6} \int \frac{\log x}{x} \, dx

Substitute

u = \log x

, so

du = \frac{dx}{x}

Then

I = \frac{1}{6} \int u \, du = \frac{1}{6} \cdot \frac{u^2}{2} + C = \frac{1}{12} (\log x)^2 + C

Rewrite in terms of

\sqrt{x}

(\log x)^2 = 4 [\log(\sqrt{x})]^2

, so

I = \frac{1}{12} \cdot 4 [\log(\sqrt{x})]^2 + C = \frac{1}{3} [\log(\sqrt{x})]^2 + C

This matches option C.

Answer:

\frac{1}{3} [\log(\sqrt{x})]^2 + C

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