1. ## Integration by Substitution

just two problems that are giving me grief

$\displaystyle \int(x^2)+\frac{1}{(3x)^2} dx$

$\displaystyle \int\frac{x^2+3x+7}{\sqrt{x}} dx$

thanks

2. For the first one:
$\displaystyle \int x^2 + \frac{1}{(3x)^2}dx$

I would just split them up to make it easier like this and also just fix up that $\displaystyle 3x^2$ to make it look a bit nicer:

$\displaystyle \int x^2dx + \int \frac{1}{9x^2}dx$

The first part is easy...

$\displaystyle \int x^2dx = \frac {x^3}{3}$

The second part is a little bit harder...

$\displaystyle \int \frac{1}{9x^2} dx = \frac{1}{9} \int x^-2dx$

Then for the $\displaystyle \int x^-2dx$ all it is... is $\displaystyle \frac {x^-1}{-1}$

So for a final answer you have...
$\displaystyle \frac {x^3}{3} - \frac{1}{9x}$

3. Originally Posted by larson
For the first one:
$\displaystyle \int x^2 + \frac{1}{(3x)^2}$

I would just split them up to make it easier like this and also just fix up that $\displaystyle 3x^2$ to make it look a bit nicer:

$\displaystyle \int x^2 + \int \frac{1}{9x^2}$

The first part is easy...

$\displaystyle \int x^2 = \frac {x^3}{3}$

The second part is a little bit harder...

$\displaystyle \int \frac{1}{9x^2} = \frac{1}{9} \int x^-2$

Then for the $\displaystyle \int x^-2$ all it is... is $\displaystyle \frac {x^-1}{-1}$

So for a final answer you have...
$\displaystyle \frac {x^3}{3} - \frac{1}{9x}$
don't forget you arbitrary constant, nor the dx at the end of the integrals. otherwise, good job

4. Originally Posted by doctorgk
$\displaystyle \int\frac{x^2+3x+7}{\sqrt{x}} dx$
for this one, note that $\displaystyle \sqrt{x} = x^{1/2}$. divide each term by it and then you just use the power rule

5. Originally Posted by Jhevon
don't forget you arbitrary constant, nor the dx at the end of the integrals. otherwise, good job
haha, whoops... sometimes i do forget those :\, but thanks, I added them on.