Calculate $\int \frac{\ln(15x^5)}{x} dx$

Question

I am having some trouble with the following integral:

$$\int \frac{\ln(15x^5)}{x}$$

I separated the top into:

$$\int \frac{5\ln(x)+ \ln(15)}{x}$$

But, then I don't know where to go from there. I tried $u$ substitution by letting $u=\ln(x)$.

\begin{align} \int 5u &+ \ln (15) \, du \\ \frac{5u^2}{2} &+ \ln(15)u \\ \frac{5(\ln(x))^2}{2}&+\ln(15)\ln(x) \end{align}

Where am I going wrong? Can someone please provide some hints?

Thanks a bunch!

Answer 1

$$\begin{align}\int \frac{\ln(15x^5)}{x}\,dx & = \int \frac{5\ln(x)+ \ln(15)}{x}\,dx \\ \\ & = 5\int \frac{\ln x}{x} dx + \int \ln(15)x^{-1} \,dx\end{align}$$

You integrated the first portion correctly: $u = \ln x \implies du = \frac 1x\,dx$. And it so happens that you got the second integral correct too. (Indeed, there is no real need to separate the summed integrand into the sum of two integrals.)

Recall that $\ln(15)$ is nothing more than a constant.

There is some simplification you can do, if desired (using properties of the logarithm function). But your integration is just fine.

We could, alternatively, start from the beginning and set $$u = \ln(15x^5) \implies du = \dfrac{5\cdot 15x^4}{15x^5} \,dx = \frac 5x\,dx$$

$$\int \frac{\ln(15x^5)}{x}\,dx = \frac 15\int u\,du$$

and then go from there.