Question

Evaluate the integral.\\ $\int \frac{\ln x^6}{x} dx$\\ $\frac{1}{\ln x^6} + C$\\ $\frac{1}{2} (\ln x)^2 + C$\\ $\frac{1}{12} (\ln x)^2 + C$\\ $\frac{1}{6} (\ln x)^2 + C$

          Evaluate the integral.\\
$\int \frac{\ln x^6}{x} dx$\\
$\frac{1}{\ln x^6} + C$\\
$\frac{1}{2} (\ln x)^2 + C$\\
$\frac{1}{12} (\ln x)^2 + C$\\
$\frac{1}{6} (\ln x)^2 + C$

Evaluate the integral.

∫(ln x^6)/(x) dx

(1)/(ln x^6) + C

(1)/(2) (ln x)^2 + C

(1)/(12) (ln x)^2 + C

(1)/(6) (ln x)^2 + C

Added by Anthony N.

Elementary and Intermediate Algebra

Alan S. Tussy, R. David Gustafson 5th Edition

Instant Answer

Solved by Expert Naresh Bagrecha

06/07/2021

Step 1

Step 1: Apply the power rule of integration, which states that the integral of x^n with respect to x is (1/(n+1))x^(n+1) + C, where C is the constant of integration. Show more…

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