Question

I am practicing integration by parts....
(watched some videos, read my textbook, and coming at it just a bit...)

Answer 1

\[\Large \int\limits_{ }^{ }xe^{3x}~dx\]\[\Large \int\limits_{ }^{ }xe^{3x}~dx=\frac{1}{3}xe^{3x}-\Large \int\limits_{ }^{ }\frac{1}{3}e^{3x}~dx\]

Answer 2

So I am differentiating the x, and by doing so I am reducing the x from the integral, is that correct?

Answer 3

Yes.

Answer 4

well, if that is correct, my final answer would be,
\[\large \int\limits_{ }^{}xe^{3x}~dx=\frac{1}{3}xe^{3x}-\frac{1}{9}e^{3x}+C\]

Answer 5

In general: \[
\int u\;dv = uv - \int v\;du
\]Typically you want to pick \(u\) to be something who's derivative is very simple to integrate, and for functions like \(e^x,\sin x,\cos x\) who's derivatives loop, y ou want to set them as \(v'\)

Answer 6

Since you said it, (the e^x and cos(x) ) I will mention that I watched the video (from Khan)
and there he integrates by parts, the e^x cos(x)

but, unlike my expectations, after doing integration by parts twice or three times (don't remember) he just solves for an integral, just like solving an algebra equation.....
you know what I mean (he didn't actually come to "integrate" the function, or integrate, as what I would generally mean by that word.

And anyway, I am going to continue doing another problem....

Answer 7

\[\large \int\limits_{ }^{ }\tan^{-1}x~dx\]so the other, the second function, is 1.
\[\large \int\limits_{ }^{ }\tan^{-1}x~dx=x \tan^{-1}x-\large \int\limits_{ }^{ }\frac{x}{1+x^2}~dx\]
taking the second integral by u substitution...
\[u=1+x^2~~~~~\frac{1}{2}du=dx\]we get,\[\large \int\limits_{ }^{ }\tan^{-1}x~dx=x \tan^{-1}x-\large \frac{1}{2}\int\limits_{ }^{ }\frac{1}{u}~du\]\[\large \int\limits_{ }^{ }\tan^{-1}x~dx=x \tan^{-1}x-\large \frac{1}{2} \ln(1+x^2)+C \]

Answer 8

When you have a case of \(\cos x,\sin x\) and \(e^x\), you want to let \(v'\) be the \(e^x\). The \(\cos x,\sin x\) will eventually return to their original form, and you will end up with the integral becoming a variable for which that you can solve.

Answer 9

yeah, that's how I thought... and how about my problem of arctan, excuse me for insisting?

Answer 10

It looks correct.

Answer 11

TY!

Answer 12

closing....