Question

Would someone be willing to check my work on taking this anti-derivative?

**I'll post the equations as the 1st comment.

Any and all help is greatly appreciated!

Answer 1

The problem:
\[f(x) = \frac{p^{6+\sqrt{p}} + 3p^4-2p^2}{p^5}\]
\[F(x) = \frac{p^{\sqrt{2}+2}}{\sqrt{2}+2} + 3\ln|p| + p^{-2} + C\]

Answer 2

f(p), yes? :)

Answer 3

Ah, yes:) Force of habit– my bad!

Answer 4

Other than that, would you say it's good?

Answer 5

I dunno, gimme a few minutes hehe ^^

Answer 6

Last two terms look good!
Hmmmm, a little confused on that first one...
Was the original problem supposed to have a sqrt(2) and not a sqrt(p) in it?

Answer 7

\(\large \color{black}{f(p) = \dfrac{p^{6+\sqrt{p}} + 3p^4-2p^2}{p^5}}\)

\(\large \color{black}{f(p) = p^1+p^{-4.5}+3p^{-1}-2p^{-3}}\)


\(\large \color{black}{F(p) = \dfrac{p^{1\color{red}{+1}}}{{1\color{red}{+1}}}+\dfrac{p^{-4.5\color{red}{+1}}}{-4.5\color{red}{+1}}+3\ln|p|+\frac{2p^{-3\color{red}{+1}}}{-3\color{red}{+1}}}\)

Answer 8

Oh, no I am wrong

Answer 9

p^(√p) ?
that has a closed form integral?

Answer 10

Yes, apparently it does :)
I'm trying to remember the technique for dealing with it hehe

Answer 11

http://www.wolframalpha.com/input/?i=x%5E%28%E2%88%9Ax%29+integral

what?

Answer 12

zepdrix, is that supposed to be like this?

Answer 13

applying the power rule, when x is in the exponent aand the base??
Maybe I don't know something about elementary rules of integration?

Answer 14

Ya they applied power rule for integration.
I'm trying to work it out on paper to understand it.

The process probably simplifies down to that rule :o
Interesting!

Answer 15

It is quite wierd... x^x would yeild some redicoulous results with gamma and other stuff, but when the power is √x we get the power rule..... let's see..

Answer 16

\(\color{blue}{\displaystyle \int x^{\sqrt{x}}dx}\)

Answer 17

Yes @zepdrix, it's supposed to be (sqrt.2) in the original function's exponent.

Answer 18

Oh lol :)
Ok that makes things simpler then.

Answer 19

√2? nice

Answer 20

Should it be 1+sqrt(2) leading to 2+sqrt(2)?
You have 6+sqrt(2) written right now I guess.

Answer 21

\(\large \color{black}{f(p) = \dfrac{p^{6+\sqrt{2}} + 3p^4-2p^2}{p^5}}\)

\(\large \color{black}{f(p) = p^1+p^{\sqrt{2}~-5}+3p^{-1}-2p^{-3}}\)


\(\large \color{black}{F(p) = \dfrac{p^{1\color{red}{+1}}}{{1\color{red}{+1}}}+\dfrac{p^{\sqrt{2}~-5\color{red}{+1}}}{\sqrt{2}~-5\color{red}{+1}}+3\ln|p|+\frac{2p^{-3\color{red}{+1}}}{-3\color{red}{+1}}}\)

Answer 22

then, simplifying is on you

Answer 23

but I will still take a look at integral of course...

Answer 24

Oh yes your work looks good amon :)
I was being silly.

I forgot that we subtract 5 from the 6.

Answer 25

Good tip for the work-around there too @zepdrix :)

Answer 26

idku,
I think you accidentally subtracted 5 `twice`.

Answer 27

Thank you!

Answer 28

and maybe did something else silly too hehe

Answer 29

it seems correct to me?
maybe I am just to tired to actually think...