Evaluate the indefinite integral 9dx/xln6x

Question

I have to evaluate the indefinite integral ∫9dx/xln6x but I'm getting stuck with the solution.

When I tried to solve it I got 9lnln6x + C which is only partly the right answer... apparently? Evidently my answer was only 50% of the correct answer, as there is a more general solution or something like that? I have no idea, what, if anything, that I'm doing wrong. What is an answer with "more than one possibility?" I am very confused by this.

Answer 1

I'm not going to go into the step-by-step detail of evaluating the indefinite integral because your answer is essentially correct. My best guess as to why your answer is only '50% correct' is because the integral of 1/x is ln|x|. Thus, the more precise indefinite integral should be:

Hope this was helpful.

Answer 2

Step 1: Substitution

You can use the substitution method to simplify this integral:

u = ln(6x)

Now, differentiate u with respect to x:

du/dx = 1/x

Therefore, dx = du / x

Substitute this into the integral:

∫ 9 dx / (x ln(6x)) = 9 ∫ du / u

Step 2: Evaluate the Integral

Now, you can easily integrate:

9 ∫ (1/u) du = 9 ln|u| + C

Step 3: Substitute u back

Recall that u = ln(6x), so substituting back, you get:

9 ln|ln(6x)| + C

Final Answer

9 ln|ln(6x)| + C

Answer 3

I have to assume you mean ∫ 9/(xln(6x)) dx

This is the same as 9 * ∫ (ln(6x))^-1 * 1/x dx

Right here, we see a function of ln(6x), multipied by the derivative of ln(6x). In general, whenever you see an integral of the form f(g(x)) * f'(x), if you can recognize it, you should be able to figure out that the answer is F(g(x)) ("F" being the integral of f).

However, to be more formal, and to be sure we're doing it right (for both your sake and your teacher's sake), let's instead use substitution.

u = ln(6x)

du = 1/x dx

9 * ∫ 1/u du

9 * ln(u) + C

u = ln(6x)

9ln(ln(6x)) + C.

Or, alternatively, 9ln|ln(6x)|+C, depending on your teacher.

Technically, the absolute value bars are only correct here if you strictly perform the integral on real numbers and never integrate between values of opposite signs. But, unless you've taken complex analysis, integrating into the complex plane probably isn't going to come up, and many teachers actually insist that you use the absolute value bars, so I guess better safe than sorry.