Integrating $\int \frac{x^2}{\sqrt{2+3x}}$ using u-sub and integration by parts I'm supposed to integrate this problem by two methods and show that they are the same or that they differ by a constant.
Problem: $$\int \frac{x^2}{\sqrt{2+3x}} dx$$
First method: u-sub with $u = 2+3x$
Second method: integration by parts $u = x^2$ and $dv = \frac{1}{\sqrt{2+3x}}$

Method 1) 
$$ u = 2 + 3x $$
$$ \frac{1}{3} du = dx$$
$$\frac{1}{3} \int \frac{x^2}{\sqrt{u}} du $$
Am I able to finish this up with tabular integration? Or do I need to do an integration by parts? The $u^\frac{-1}{2}$ seems to be where it will likely get complicated which is why I ask about tabular integration.

Method 2)
$$ u = x^2$$
$$ \frac{1}{2} du = dx $$
$$ dv = \frac{1}{\sqrt{2+3x}} $$
Now for v, it looks like I'll have to u-sub...
$$ w = 2+3x $$
$$ \frac{1}{3} dw = dx $$ 
$$ v = \frac{1}{3} \int \frac{dw}{w} $$
$$ v = \frac{1}{3} ln | 2+3x | $$
Now we put it together with integration by parts
$$ (x^2)(\frac{1}{3} ln |2+3x| - \int ln |2+3x| (\frac{1}{2}) $$
Now I'll have to integrate the second part
$$ \frac{1}{2} ln |2 + 3x | $$ 
u-sub
$$ u = 2+3x $$
$$ \frac{1}{3} du = dx $$
New equation is
$$ \frac{1}{6} \int \ln | u | du $$
Integration by parts
$$ w = u $$
$$ dw = \frac{1}{u} $$ 
$$ dv = du $$
$$ v = u $$
So that'll give us
$$ \frac{1}{6} [w*w - \int u*\frac{1}{w}] dw$$
which is 
$$ \frac{1}{6} [w^2 - \int \frac{w}{w}] dw$$
reduce the w and integrate it
$$ \frac{1}{6} [ w^2 - w ] dw $$
Now we need to plug in our u for w. ($w^2 = \sqrt{u}$)
$$ \frac{1}{6} [ \sqrt{u} - u ] du $$
Now lets plug in 2 + 3x for our u
$$ \frac{1}{6} [ \sqrt{2+3x} - 2+3x $$
Put our integrals together...
$$ (x^2)(\frac{1}{3} ln |2+3x| - \frac{1}{6} [ \sqrt{2+3x} - 2+3x ] + c $$

I'm not sure if I did this correctly so I would appreciate it if anyone can check it and also hopefully advise me on the first method as well.

Edited Answers...
Method 1:
$$\frac{1}{27} [ \frac{2}{5}(2+3x)^{5/3} - \frac{8}{3}(2+3x)^{3/2} - 8(2+3x)^{1/2}] + c$$
Method 2:
$$ x^2 \sqrt{2+3x} - \frac{2}{3} [ 2x(\frac{2}{3}(2+3x)^{3/2}) - \frac{1}{3}(\frac{2}{5}(2+3x)^{5/2} ] + c $$
 A: In the first case, you need to express $x^2$ in terms of $u$, by noting $$u = 2 + 3x \iff x =\frac 13(u -2).$$ So $$x^2 = \dfrac 19 (u - 2)^2$$
In the second case, $u = x^2 \implies du = 2x dx \iff \dfrac 12 u\,du = x\,dx$
Also, for $dv = \dfrac 1{\sqrt{2 + 3x}}$, and $w = 2 + 3x \implies \dfrac 13 dw = dx$, then $$dv = \frac 1{\sqrt w}\cdot \,dw \implies v = \frac 13 \int w^{-1/2} \,dw = \dfrac{{\frac 13 w^{\frac 12}}}{\frac 12} + C = \dfrac 23w^\frac 12 + C = \frac 23\sqrt{2 + 3x} + C$$
A: You are going to love how simple the solution is.

Method 1 u sub
U=(3x+2) x=1/3(u-2)
DX=1/3(du)
Integrand
(1/27)(u-2)^(2)*(u)^(1/2)
Here is the easy way out!
Expand the square of the binomial and then multiply the root u throughout!
I know u r saying 'why didn't I think of that!' And voila! You have solved the first part!
Now to the second part IBP!
Here also you have to kiss the answer! Kip it simple senor!
Let u=x^2.   And. DV=(3x+2)^(1/2)DX
Our answer will be
uv-int(v du)
But vdu looks like
(3x+2)^(3/2)* x DX
Again I will take the easy out!
x (3x+2) (3x+2)^(1/2).  I know u r saying its not getting any easier!
Do a u sub! u=(3x+2) solve for x! Plug and chug above! And multiply through!
And voila you are on home base! Without a sweat Mr babe Ruth!
Wishmath stack had a "equation editor" hope you understood the unformatted math solution!
Always train your eye and not your mind for the easy way out!
Enjoy!
