So here is what I did first.

$$∫16\ln(x^{1/3})dx$$ move the constant $16$ out


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


$$u=\ln(x^{1/3})\qquad dv=∫16 dx du=\frac{1}{x^{1/3}}\frac{1}{3x^{2/3}} dx\qquad v= 16x$$

simplifies to


$$16x \ln(x^{1/3}-∫16x\frac{dx}{3x}$$

$\frac{16x}{3x}$ 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


and this worked I want to know why $\frac{16}{3}$ can't cross the integral sign.

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    $\begingroup$ Please use $\LaTeX$ in your posts. This is too hard to read. $\endgroup$ – Mnifldz Jan 30 '15 at 9:02
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    $\begingroup$ What makes you think your first method wasn't correct? It is correct once you place your parentheses correctly: $(16/3) (x\ln x-x)+C$. $\endgroup$ – David Mitra Jan 30 '15 at 9:05
  • $\begingroup$ @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. $\endgroup$ – Lefty Jan 30 '15 at 9:18

Your method is correct. Observe that

\begin{eqnarray*} \frac{d}{dx} \left [\frac{16}{3} (x\ln x - x) + c\right ] & = & \frac{16}{3}\left (\ln x + \frac{1}{x}\cdot x - 1 \right) \\ & = & \frac{16}{3} \ln x \\ & = & 16 \ln x^{1/3}. \end{eqnarray*}

In general you can check an integral computation by taking the derivative of your result and seeing if you obtain what you started with.


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