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

$$ \frac{1}{2} ln |2 + 3x | $$


$$ u = 2+3x $$ $$ \frac{1}{3} du = dx $$

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:

$$ x^2 \sqrt{2+3x} - \frac{2}{3} [ 2x(\frac{2}{3}(2+3x)^{3/2}) - \frac{1}{3}(\frac{2}{5}(2+3x)^{5/2} ] + c $$

  • $\begingroup$ I have a few words of advice: DON'T write $x$ and $u$ (or $du$) in the same integral. It may not necessarily be mathematically incorrect, but when I see students do it, they NEVER get the right answer. Second, when I see students use the tabular method of integration for simple integration by parts problems, instead of the standard method, they usually get the wrong answer. $\endgroup$ – Stefan Smith Sep 27 '13 at 23:54

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 $$dv = \frac 1{\sqrt w}\cdot \,dw \implies v = \frac 13 \int w^{-1/2} \,dw = \dfrac{{\frac 13 w^{\frac 12}}}{\frac 12} + C = \dfrac 23w^\frac 12 + C = \frac 23\sqrt{2 + 3x} + C$$

  • $\begingroup$ Will work on the first method with the new info! For the second method, isn't that what I have? $\endgroup$ – ConfusingCalc Sep 27 '13 at 18:44
  • $\begingroup$ You had :$\frac 12 du = dx,$ and you should have $\frac 12 du = x\,dx$. Also see my update $\endgroup$ – Namaste Sep 27 '13 at 18:56
  • $\begingroup$ Man, this is a difficult one. With the first case: $\frac{1}{27} \int (u-2)^2 * (u)^{-1/2} du$ do I do a u-sub again or go directly into IBP? $\endgroup$ – ConfusingCalc Sep 27 '13 at 19:15
  • $\begingroup$ Looks good. You can expand out the binomial, multiply through by $u^{-1/2},$ and integrate using the power rule! Think about accepting an answer...I'll keep checking in to follow up!) $\endgroup$ – Namaste Sep 27 '13 at 20:00
  • $\begingroup$ Unfortunately since this is such a massive question, I'm lost on where I've messed up. I tried expanding the binomial in prackash's answer, but I'm not sure if that is right. Sorry, slightly confused with all the text at the moment. $\endgroup$ – ConfusingCalc Sep 27 '13 at 20:06
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!

  • $\begingroup$ Working method 1 the way you mention gives me: $$\frac{1}{27} ( \frac{8x-6}{\sqrt{2+3x}}) + c$$ are you sure that is correct? $\endgroup$ – ConfusingCalc Sep 27 '13 at 19:55
  • $\begingroup$ By the way, I quite enjoyed your reply. Good advice and a nice sense of humor. The one thing I noticed is that it should be $u^{-1/2}$ rather than $u^{1/2}$ $\endgroup$ – ConfusingCalc Sep 27 '13 at 20:23

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