Unspecific integration for a square root

[YehiaMedhat]

I actually have solved it by diffrentiating the answers, but what would be a systematic way to solve it.
I began tried to [math]u = \sqrt (3-2x-x^2)[/math], but this ended up having the next [math]\int \frac {du}{-1-x}[/math], as i remeber. So what do you think?

 

[BigBeachBanana]

View attachment 34650
I actually have solved it by diffrentiating the answers, but what would be a systematic way to solve it.
I began tried to [math]u = \sqrt (3-2x-x^2)[/math], but this ended up having the next [math]\int \frac {du}{-1-x}[/math], as i remeber. So what do you think?

Hint:
1) Complete the square under the radical.
2) Recognize that it's an integral of Arcsin(x) by making a substitution

 

[Dr.Peterson]

View attachment 34650
I actually have solved it by diffrentiating the answers, but what would be a systematic way to solve it.
I began tried to [math]u = \sqrt (3-2x-x^2)[/math], but this ended up having the next [math]\int \frac {du}{-1-x}[/math], as i remeber. So what do you think?

First, the word you want is not "unspecific" but "indefinite". A dictionary would call them synonyms, but the latter is used in this context.

As for integrating it, I would start by completing the square in the radicand. Then you'll be able to do a substitution.

Also, please show us the details of your work, so we can correct specific errors you might be making.

 

[YehiaMedhat]

Well, thanks?