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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?

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

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?

  1. $\int t^2 \sin (\beta t) dt$

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

  1. $\int \ln \sqrt{x} dx$

$$\int \ln \sqrt{x} dx = \int \ln x^{\frac{1}{2}} dx$$

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?

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    $\begingroup$ You pick $u$ and $dv$ in order to make $\int v\,du$ easier to calculate than $\int u\,dv$. So you shouldn’t pick $dv$ to be something you can’t integrate, since you need $v$. And you should generally pick $u$ to be something whose derivative $du$ is “simpler.” Use these ideas to think about (2) and (3). $\endgroup$ Jan 30 '19 at 5:14
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    $\begingroup$ Note $\log(x^{1/2})=\frac12\log(x)$. Then let $u=\log(x)$ and $v=x$. $\endgroup$
    – Mark Viola
    Jan 30 '19 at 5:14
  • $\begingroup$ math.stackexchange.com/questions/768332/liate-how-does-it-work $\endgroup$ Jan 30 '19 at 5:21
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  1. You seemed to have missed out a minus sign on the integral on line 2.

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

  1. 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.

  2. 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.

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  • $\begingroup$ How did you see this for 2? Can you show me? $\endgroup$
    – Jwan622
    Jan 30 '19 at 13:43
  • $\begingroup$ Something in the form of $\int x^n \sin(mx)dx$ or $\int x^n \cos(mx)dx$ is a fairly common question to test how well you know integration by parts. $\endgroup$
    – Infiaria
    Jan 30 '19 at 13:52
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  1. 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.

  1. 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):

$$\int f(x) g'(x)dx=f(x)g(x)-\int f'(x)g(x)dx$$

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  • $\begingroup$ Regarding number 1... in wolfram's example doesn't the -1/9 distribute to both parts inside the parenthesis? Isn't it identical to mine then? $\endgroup$
    – Jwan622
    Jan 30 '19 at 13:55
  • $\begingroup$ In that respect, yes, it is identical. But you have a problem with the minus sign. $\endgroup$ Jan 30 '19 at 13:58
  • $\begingroup$ Isn't it supposed to be (-1/9) in the answer? $\endgroup$
    – Jwan622
    Jan 30 '19 at 14:04
  • $\begingroup$ Yes, see my answer. I edited it a little bit. There is a $-1/9$ in front of the rest of the answer. Can you see it? $\endgroup$ Jan 30 '19 at 14:06
  • $\begingroup$ Nope? What do you mean? $\endgroup$
    – Jwan622
    Jan 30 '19 at 21:37

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