Question

Find the anti-derivative of x(x^2+1)^-1/2

Answer 1

\[\frac{ x }{ \sqrt{x^2+1} }\]

Answer 2

set u=x^2+1

Answer 3

I will give you a latex tip.
`\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x}{ \sqrt{x^2+1} }~dx}\)` this I use, to put up:
\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x}{ \sqrt{x^2+1} }~dx}\)

Answer 4

but, really, try \(\large\color{slate}{\displaystyle u=x^2+1}\)

Answer 5

what will you get for du ?

Answer 6

2x

Answer 7

yes, so \(\large\color{slate}{\displaystyle du=2x~dx}\)

Answer 8

you had: \(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x}{ \sqrt{x^2+1} }~dx}\). Now do you substitution.

Answer 9

I didn't think you could find the derivative of each individual part?....

Answer 10

you can't

Answer 11

So you're saying... \[\frac{ x }{ \sqrt{u} }\] ?

Answer 12

no

Answer 13

there are two ways to write this.

way 1

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x}{\sqrt{x^2+1}}dx}\)

\(\large\color{gray}{\displaystyle u=x^2+1}\)
\(\large\color{gray}{\displaystyle du=2x~dx~~~~~\Longrightarrow~~~~~dx=\frac{du}{2x}}\)

\(\large\color{black}{\displaystyle\int\limits_{~}^{~}\frac{x}{\sqrt{u}}\frac{du}{2x}}\)

the x's cancel and you take the 1/2 constant out.

\(\large\color{black}{\displaystyle \frac{1}{2}\int\limits_{~}^{~}\frac{1}{\sqrt{u}}du}\)

Answer 14

I keep disconnecting sorry

Answer 15

All good. I'm still soaking in the last response. Trying to comprehend.

Answer 16

way 2.

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{x}{\sqrt{x^2+1}}~dx}\)

\(\large\color{red}{\displaystyle u=x^2+1}\)
\(\large\color{slate}{\displaystyle du=2x~dx ~~~~~~~\Rightarrow}\) \(\large\color{green}{\displaystyle \frac{1}{2}du=x~dx }\)

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\color{green}{x }}{\sqrt{\color{red}{\displaystyle x^2+1}}}~\color{green}{dx }}\)

substitution,

\(\large\color{slate}{\displaystyle\int\limits_{~}^{~}\frac{\color{green}{\frac{1}{2} }}{\sqrt{\color{red}{\displaystyle u}}}~\color{green}{du }}\)

take the constant out,

\(\large\color{slate}{\displaystyle\color{green}{\frac{1}{2}}\int\limits_{~}^{~}\frac{\color{green}{1 }}{\sqrt{\color{red}{\displaystyle u}}}~\color{green}{du }}\)

Answer 17

you get the exact same thing, but a substitution is conceptualized a little differently.

Answer 18

I think 1st approach is more standard though, so use that one I would advise.

Answer 19

Right... so now I would take the anti-derivative of 1/sqrtu in terms of u?
The second example is easier to understand.. thank you.

Answer 20

Why is du = 2xdx?

Answer 21

oh, if you want 2nd approach, then it would also work.....

And then the anti-derivative of u is not hard.
re-write it as u^(-1/2)

Answer 22

Nevermind. Got it.

Answer 23

oh, ok.

Answer 24

Okay... anti-derivative of u^-1/2 would be 2sqrtu

Answer 25

No. The two would cancel?

Answer 26

So the final answer is \[\sqrt{x^2+1}\]

Answer 27

yes, the expression is good. But you left out the +c

Answer 28

Darn. I always do that. Lol

Answer 29

Thank you so much!!

Answer 30

\(\large\color{slate}{\displaystyle \frac{d}{dx}~(x^2+1)^{1/2}+C= \\ ~\\~\\ \frac{1}{2}(x^2+1)^{1/2~~-1~~}\color{blue}{\times~2x}\\ ~\\~\\ \frac{1}{2}(x^2+1)^{-1/2}\color{blue}{\times~2x}\\ ~\\~\\ (x^2+1)^{-1/2}\color{blue}{\times~x}~\\~\\ \displaystyle\frac{x}{\sqrt{x^2+1}}}\)

Answer 31

yes, just checked...

Answer 32

(yes, the blue is the chain rule)

Answer 33

I did the same. Looks great. Thanks again.

Answer 34

sure:)