Integration by parts questions. Work check. I have a couple of problems that I'm trying to work through. I'm a tad stuck on 2. Here is what I have?

  
*
  
*$\int t \cdot e^{-3t} dt$

so let's say:
$$u = t \quad \text{and} \quad du = dt$$
$$dv = e^{-3t} \quad \text{and} \quad v = \frac{e^{-3t}}{-3}$$
so according to integration by parts:
$$\begin{align*} \int t \cdot e^{-3t} dt &= t \cdot \frac{e^{-3t}}{-3} - \int \frac{e^{-3t}}{-3} dt \newline &= \frac{t}{-3} \cdot e^{-3t} - \frac{-1}{3} \frac{e^{-3t}}{-3} \newline &= \left(\frac{t}{3} - \frac{1}{9} \right) e^{-3t} \end{align*}$$
Is this right? 


  
*$\int t^2 \sin (\beta t) dt$

Is $\beta $ a constant? What is this notation?


  
*$\int \ln \sqrt{x} dx$

$$\int \ln \sqrt{x} dx = \int \ln x^{\frac{1}{2}} dx$$
so let's try:
$$u = \ln{x^{\frac{1}{2}}} \quad \text{and} \quad du = \frac{1}{x^{\frac{1}{2}}} \cdot \frac{1}{2} \frac{1}{x^{\frac{1}{2}}} = \frac{1}{2x}$$
and
$$dv = dx \quad \text{and} \quad v = x$$
so
$$\begin{align*} \int \ln \sqrt{x} dx = \ln x^{\frac{1}{2}} \cdot x - \int x dx = x \ln{x^{\frac{1}{2}}} - \frac{x^2}{2} \end{align*}$$
How does that look?
 A: *

*You seemed to have missed out a minus sign on the integral on line 2.


$$- \int \frac{e^{-3t}}{-3} dt = - \frac{1}{-3}\frac{e^{-3t}}{-3} + c$$


*If it doesn't state it in the question, then yes, $\beta$ is a constant. As long as it doesn't say that $\beta$ is some function of $t$, since the integral is with respect to $t$. 
For this question you need to integrate by parts twice, choosing to differentiate the $t$ term every time.

*You seemed to have substituted the variables wrong.
$$= x\ln x^{\frac{1}{2}} - \int \frac{x}{2x} dx$$
Remember that $\int k\cdot f(x) dx$, where $k$ is a constant is equal to $k\int f(x) dx$. In this case realise that $\log x^n = n\log x$. This should make it a bit less messy.
A: *

*I'm going to check your work by integrating it too, but I'm' going to do it my way. Here's my solution:


$$
\begin{align}
\int xe^{-3x} dx
&=-\frac{1}{3}\int x\left(e^{-3x}\right)' dx\\
&=-\frac{1}{3}\left(xe^{-3x}-\int e^{-3x} x'dx\right)\\
&=-\frac{1}{3}\left(xe^{-3x}-\int e^{-3x} dx\right)\\
&=-\frac{1}{3}\left(xe^{-3x}-\left(-\frac{1}{3}\right)\cdot\int\left(e^{-3x}\right)'dx\right)\\
&=-\frac{1}{3}\left(xe^{-3x}+\frac{1}{3}e^{-3x}\right)\\
&=-\frac{1}{3e^{3x}}\left(x+\frac{1}{3}\right)\\
&=-\frac{1}{3e^{3x}}\cdot\frac{1}{3}\cdot\left(3x+1\right)\\
&=-\frac{3x+1}{9e^{3x}}+C.
\end{align}
$$
WolframAlpha's solution is identical to mine: link. Your answer is not correct. And please remember that there always has to be a constant of integration, $C$, at the end of your answer.


*Yes, it's a constant, but I don't know what you mean by "What is this notation?".


3.
$$
\begin{align}
\int \ln \left(\sqrt{x}\right) dx
&=\int \ln\left(\sqrt{x}\right) x'dx\\
&=x\ln\left(\sqrt{x}\right)-\int x \left[\ln\left(\sqrt{x}\right)\right]' dx\\
&=x\ln\left(\sqrt{x}\right)-\int x \frac{1}{\sqrt{x}}\frac{1}{2\sqrt{x}} dx\\
&=x\ln\left(\sqrt{x}\right)-\frac{1}{2}\int dx\\
&=x\ln\left(\sqrt{x}\right)-\frac{x}{2}\\
&=x\left(\ln\left(\sqrt{x}\right)-\frac{1}{2}\right)+C.
\end{align}
$$
To do integration by parts, I just directly used this formula which is equivalent to the other formula you're using (the one with $v$'s and $u$'s):
$$\int f(x) g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$$
