So here is what I did first.
$$∫16\ln(x^{1/3})dx$$ move the constant $16$ out
$$16∫\ln(x^{1/3})dx$$
use properties of logarithms to rewrite natural log of cube root of $x$ as $\ln x$ divided by $3$ and move out $1/3$
$$\frac{16}{3}∫\ln x dx$$
integration by parts: $$u=\ln x\qquad dv=1dx\\ du=\frac{1}{x}dx\qquad v=x$$
$$\frac{16}{3}x\ln x-∫\frac{x dx}{x}$$
$$\frac{16}{3}x\ln x-∫dx$$
$$\left(\left(\frac{16}{3}x\ln x\right)-x\right)+c$$
That wasn't correct so i tried leaving the constant 16 inside thinking i can use that as my dv and did the following
$$∫16\ln(x^{1/3})dx$$
$$u=\ln(x^{1/3})\qquad dv=∫16 dx du=\frac{1}{x^{1/3}}\frac{1}{3x^{2/3}} dx\qquad v= 16x$$
simplifies to
$$du=\frac{dx}{3x}$$
$$16x \ln(x^{1/3}-∫16x\frac{dx}{3x}$$
$\frac{16x}{3x}$ cancels to $\frac{16}{3}$, left with $\frac{16}{3}∫1dx$ where the integral is just $x$
$$16x \ln(x^{1/3})-\frac{16x}{3}+C$$
further simplified
$$16x\ln(x^{1/3})-\frac{1}{3}+C$$
and this worked I want to know why $\frac{16}{3}$ can't cross the integral sign.
