• Sep 21st 2013, 05:46 PM
wondering
$\int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}$

I let $u=^3\sqrt(x)-1$

$u+1=^3\sqrt(x)$

$\int\frac{u+1}{u}du$
$\int (1 + \frac{1}{u})du$
$u + ln\abs{u} + c$

$^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c$

Am I close? Wolframalpha comes up with a different method. Thanks.
How do I get the cube root in Latex?
• Sep 21st 2013, 06:24 PM
FelixFelicis28
Quote:

Originally Posted by wondering
$\int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}$

I let $u=^3\sqrt(x)-1$

$u+1=^3\sqrt(x)$

$\int\frac{u+1}{u}du$
$\int (1 + \frac{1}{u})du$
$u + ln\abs{u} + c$

$^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c$

Am I close? Wolframalpha comes up with a different method. Thanks.
How do I get the cube root in Latex?

You have forgetten to make a substitution for $dx$.

$u = x^{1/3} - 1 \implies \frac{du}{dx} = 1/3 x^{-2/3} \implies dx = du 3(u+1)^2$

Btw, the LaTeX for the cube root is $\sqrt[3]{x}$
• Sep 21st 2013, 06:27 PM
HallsofIvy
Quote:

Originally Posted by wondering
$\int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}$

I let $u=^3\sqrt(x)-1$

$u+1=^3\sqrt(x)$

$\int\frac{u+1}{u}du$

Okay, if $u= ^3\sqrt{x}- 1$ then $u+ 1= ^3\sqrt{x}$ and so $\frac{^3\sqrt{x}}{^3\sqrt{x}- 1}= \frac{u+ 1}{u}$ but what happened to the "dx"? if $^3\sqrt{x}= u+ 1$ then $x= (u+ 1)^3= u^3+ 3u^2+ 3u+ 1$ so [tex]dx= (3u^2+ 6u+ 3)du
$\int \frac{^3\sqrt{x}}{^3\sqrt{x}- 1}dx= \int \frac{u+1}{u}(3u^2+ 6u+ 3)du= \int (3u^2+ 6u+ 3+ 3u+ 6+ \frac{3}{u})du= \int (3u^2+ 9u+ 9+ \frac{3}{u})du$

Quote:

$\int (1 + \frac{1}{u})du$
$u + ln\abs{u} + c$

$^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c$

Am I close? Wolframalpha comes up with a different method. Thanks.
How do I get the cube root in Latex?
The best way to get a cube root (though I dislike it my self!) is "\sqrt[3]{x}" gives $\sqrt[3]{x}$. I have enough trouble with people who mistake $\sqrt[3]{3}$ for a square root without Latex reinforcing it! Your "^3\sqrt{x}" is a good substitute.
• Sep 21st 2013, 06:29 PM
FelixFelicis28
Quote:

Originally Posted by HallsofIvy
What you did is the best way to get a cube root (though I dislike it my self!) "^3\sqrt{x}" gives $^3\sqrt{x}$.

\sqrt[n]{x} $=\sqrt[n]{x}$

:D
• Sep 21st 2013, 06:33 PM
HallsofIvy
Darn! I was hoping to edit my post so I wouldn't look so foolish but I wasn't fast enough!
• Sep 21st 2013, 06:43 PM
FelixFelicis28
Just too quick. ;)
• Sep 21st 2013, 07:15 PM
Prove It
Quote:

Originally Posted by wondering
$\int \frac{^3\sqrt(x)}{^3\sqrt(x)-1}$

I let $u=^3\sqrt(x)-1$

$u+1=^3\sqrt(x)$

$\int\frac{u+1}{u}du$
$\int (1 + \frac{1}{u})du$
$u + ln\abs{u} + c$

$^3\sqrt(x) -1 + ln\abs{^3\sqrt(x)} + c$

Am I close? Wolframalpha comes up with a different method. Thanks.
How do I get the cube root in Latex?

\displaystyle \begin{align*} \int{\frac{\sqrt[3]{x}}{\sqrt[3]{x} - 1}\,dx} &= \int{\frac{\left( \sqrt[3]{x} \right) ^2 \sqrt[3]{x} }{\left( \sqrt[3]{x} \right) ^2 \left( \sqrt[3]{x} - 1 \right) } \, dx} \\ &= 3\int{ \frac{1}{3\left( \sqrt[3]{x} \right) ^2} \cdot \frac{x}{\sqrt[3]{x} - 1} \, dx} \end{align*}

Now make the substitution \displaystyle \begin{align*} u = \sqrt[3]{x} - 1 \implies du = \frac{1}{3\left( \sqrt[3]{x} \right) ^2 } \, dx \end{align*} and note that this substitution also means \displaystyle \begin{align*} x = \left( u + 1 \right) ^3 \end{align*}, and the integral becomes

\displaystyle \begin{align*} 3\int{ \frac{1}{3\left( \sqrt[3]{x} \right) ^2} \cdot \frac{x}{\sqrt[3]{x} - 1} \, dx} &= 3 \int{ \frac{\left( u + 1 \right) ^3 }{u} \, du} \\ &= 3 \int{ \frac{u^3 + 3u^2 + 3u + 1}{u}\,du} \\ &= 3\int{ u^2 + 3u + 3 + \frac{1}{u}\, du} \\ &= 3 \left( \frac{u^3}{3} + \frac{3u^2}{2} + 3u + \ln{|u|} \right) + C \\ &= u^3 + \frac{9u^2}{2} + 9u + 3\ln{|u|} + C \\ &= \left( \sqrt[3]{x} - 1 \right) ^3 + \frac{9 \left( \sqrt[3]{x} - 1 \right) ^2}{2} + 9 \left( \sqrt[3]{x} - 1 \right) + 3\ln{ \left| \sqrt[3]{x} - 1 \right| } + C \end{align*}
• Sep 22nd 2013, 05:44 AM
topsquark