Integration by Substitution

**doctorgk** · February 27th 2008, 04:41 PM

just two problems that are giving me grief

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

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

thanks

**larson** · February 27th 2008, 04:56 PM

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

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

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

The first part is easy...

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

The second part is a little bit harder...

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

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

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

**Jhevon** · February 27th 2008, 05:01 PM

Originally Posted by larson

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

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

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

The first part is easy...

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

The second part is a little bit harder...

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

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

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

don't forget you arbitrary constant, nor the dx at the end of the integrals. otherwise, good job

**Jhevon** · February 27th 2008, 05:03 PM

Originally Posted by doctorgk

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

for this one, note that $\sqrt{x} = x^{1/2}$ . divide each term by it and then you just use the power rule

**larson** · February 27th 2008, 05:14 PM

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.

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