# Math Help - integration using cute root

1. ## integration using cute root

hey guys, im interested in another integration question using cute roots

integrate:

$

\int \frac{1}{\sqrt[3]{4-5x}}

$

any ideas? should i make that cute root in the form $(4-5x)^{\frac{1}{3}}$ first?

2. Hi,

Originally Posted by Title
integration using cute root
Cute root or cube root ?
Originally Posted by jvignacio
should i make that cute root in the form $(4-5x)^{\frac{1}{3}}$ first?
Yes, then you can try using $\int u' u^\alpha \,\mathrm{d}x = \frac{u^{\alpha+1}}{\alpha +1}+C$ for $\alpha \neq -1$.

3. Originally Posted by flyingsquirrel
Hi,

Cute root or cube root ?

Yes, then you can try using $\int u' u^\alpha \,\mathrm{d}u = \frac{u^{\alpha+1}}{\alpha +1}+C$ for $\alpha \neq -1$.
hhaha yes sorry cube root!!!
im not quite sure what you mean in that formular?

4. Originally Posted by jvignacio
im not quite sure what you mean in that formular?
One has $\frac{1}{\sqrt[3]{4-5x}}=\frac{1}{(4-5x)^{1/3}}=(4-5x)^{-1/3}$ so $\int \frac{\mathrm{d}x}{\sqrt[3]{4-5x}}=\int (4-5x)^{-1/3}\,\mathrm{d}x$. It looks like $\int u'u^\alpha \,\mathrm{d}x$ with $u=4-5x$ and $\alpha=-1/3$. The only term which is missing is $u'$... but we can make it appear. $u'$ is given by

$u'=\frac{\mathrm{d}(4-5x)}{\mathrm{d}x}=-5$

so if we write

$\int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x=-5\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x$

we're done since we know that $\int u'u^\alpha \,\mathrm{d}x=\frac{u^{\alpha+1}}{\alpha+1}+C$ for $\alpha\neq -1$.

5. Originally Posted by flyingsquirrel
One has $\frac{1}{\sqrt[3]{4-5x}}=\frac{1}{(4-5x)^{1/3}}=(4-5x)^{-1/3}$ so $\int \frac{\mathrm{d}x}{\sqrt[3]{4-5x}}=\int (4-5x)^{-1/3}\,\mathrm{d}x$. It looks like $\int u'u^\alpha \,\mathrm{d}x$ with $u=4-5x$ and $\alpha=-1/3$. The only term which is missing is $u'$... but we can make it appear. $u'$ is given by

$u'=\frac{\mathrm{d}(4-5x)}{\mathrm{d}x}=-5$

so if we write

$\int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x=-5\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x$

we're done since we know that $\int u'u^\alpha \,\mathrm{d}x=\frac{u^{\alpha+1}}{\alpha+1}+C$ for $\alpha\neq -1$.
ahh yes now i understand.

does the answer end up being

$
\frac {3(4-5x)^{\frac{2}{3}}}{2} + C
$

?

6. Originally Posted by flyingsquirrel
$\int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x=-5\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x$
I meant

$\int (4-5x)^{-1/3}\,\mathrm{d}x=\int \frac{(-5)}{(-5)}\times(4-5x)^{-1/3}\,\mathrm{d}x={\color{red}-\frac15}\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x$

...

Originally Posted by jvignacio
does the answer end up being

$
\frac {3(4-5x)^{\frac{2}{3}}}{2} + C
$

?
Check it by yourself . The derivative of $
x\mapsto\frac {3(4-5x)^{\frac{2}{3}}}{2} + C
$
is $
x \mapsto -5(4-5x)^{-\frac{1}{3}}
$
and it should be $x\mapsto(4-5x)^{-\frac{1}{3}}$. It turns out that you forgot the red thing :

$\int (4-5x)^{-1/3}\,\mathrm{d}x=\ldots={\color{red}-\frac15}\int \underbrace{(-5)}_{u'}\times \underbrace{(4-5x)^{-1/3}}_{u^\alpha}\,\mathrm{d}x$