1
$\begingroup$

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.

$\endgroup$
3
  • 3
    $\begingroup$ Please use $\LaTeX$ in your posts. This is too hard to read. $\endgroup$ – Mnifldz Jan 30 '15 at 9:02
  • 1
    $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.