Question

I really do not understand this Question. Please help.
Giving a medal for help.

Answer 1

\[\int\limits \frac{ x^2-3x+2 }{ x+1 } dx\]

Answer 2

\(\color{#000000 }{ \displaystyle\int\limits \frac{x^2-3x+2}{x+1}dx }\)

\(\color{#000000 }{ \displaystyle\int\limits \frac{x^2+2x+1}{x+1}+\frac{-5x+1}{x+1}dx }\)

\(\color{#000000 }{ \displaystyle\int\limits \frac{x^2+2x+1}{x+1}+\frac{-5x-5}{x+1}+\frac{+6}{x+1}dx }\)

Answer 3

this is one way to do it.
Each term is easy to simplify and then integrate.

Answer 4

Or, you can just do, u=x+1, and that would also work.

Answer 5

I am confused as to how you got those numbers.

Answer 6

solomon: would you mind explaining how you came to the conclusion that this separation of the original integrand would be proper? Remember, starting with the most basic approach is usually the most effective.

Answer 7

I agree with mathmale

Answer 8

In my mind the simplest approach would be to divide the denominator into the numerator. This is called for because the degree of the numerator is greater than that of the denominator.

Answer 9

I am assuming that you can more than comfortably factor and all that... So, what I did was that I separated each term so that you can cancel x+1, except the last term, where the rules of integration can be applied to (x+1), as if it were just any (single) variable.

Answer 10

Sid: Would you please try "long division" first: Divide (x+1) into

Answer 11

Sid: Would you please try "long division" first: Divide (x+1) into x^2-3x+2.

Answer 12

Oh Yea. I got it by doing long division. Thank you

Answer 13

So you're done? Great!

Answer 14

Yea.Thank you very much

Answer 15

My great pleasure.