# Why doesn't this method for integration by parts work?

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.

• Please use $\LaTeX$ in your posts. This is too hard to read. – Mnifldz Jan 30 '15 at 9:02
• What makes you think your first method wasn't correct? It is correct once you place your parentheses correctly: $(16/3) (x\ln x-x)+C$. – David Mitra Jan 30 '15 at 9:05
• @DavidMitra see, well I typed in my first solution and it wouldn't accept it as a correct answer i figured it was wrong. I've been working for a couple of hours now, I guess it should take a break, thank you. – Lefty Jan 30 '15 at 9:18