# Thread: integral (sqr{3}x + sqr(x)) / (sqr{6}x + 1)

1. ## integral (sqr{3}x + sqr(x)) / (sqr{6}x + 1)

$\int \frac{\sqrt[3]{x} + \sqrt{x}}{\sqrt[6]{x} + 1} dx$

and also

$\int \frac{x+8}{(x-2)(x^2+1)} dx$

2. Originally Posted by lainmaster
$\int \frac{\sqrt[3]{x} + \sqrt{x}}{\sqrt[6]{x} + 1} dx$
$\frac{\sqrt[3]x+\sqrt x}{\sqrt[6]x+1}$

$=\frac{\left(\sqrt[3]x+\sqrt x\right)\left(\sqrt[6]x-1\right)}{\left(\sqrt[6]x+1\right)\left(\sqrt[6]x-1\right)}$

$=\frac{x^{1/2}-x^{1/3}+x^{2/3}-x^{1/2}}{x^{1/3}-1}$

$=\frac{x^{2/3}-x^{1/3}}{x^{1/3}-1}$

$=\frac{x^{1/3}\left(x^{1/3}-1\right)}{x^{1/3}-1}$

$=x^{1/3}$

Now integrate.

$\int \frac{x+8}{(x-2)(x^2+1)} dx$
Use partial fraction decomposition.

Edit: And make sure you are familiar with the integral $\int\!\frac{dx}{1+x^2}.$

3. Originally Posted by Reckoner
$\frac{\sqrt[3]x+\sqrt x}{\sqrt[6]x+1}$

$=\frac{\left(\sqrt[3]x+\sqrt x\right)\left(\sqrt[6]x-1\right)}{\left(\sqrt[6]x+1\right)\left(\sqrt[6]x-1\right)}$

$=\frac{x^{1/2}-x^{1/3}+x^{2/3}-x^{1/2}}{x^{1/3}-1}$

$=\frac{x^{2/3}-x^{1/3}}{x^{1/3}-1}$

$=\frac{x^{1/3}\left(x^{1/3}-1\right)}{x^{1/3}-1}$

$=x^{1/3}$

Now integrate.
very nice!

using your final answer as motivation, we could have just factored out $x^{1/3}$ from the numerator in the beginning. then everything else cancels in the next step. but seeing that that was a good move in the first place is probably hard, i guess.

4. Thanks, Reckoner

EDIT

It seems so simple now... but how did you come up with that so easily? Is there a method, or you just saw it crystal clear from so much practice?