Long division with variables, integrate

[Ian McPherson]

the problem is integral (3x²-2x+1) / (x-2) dx. Well i set up the long division and got 3x + 4 with 9 left over, which gives me 3x+4 + 9/(x-2). I know that i need to find the integral of that, but i don't know what to do with the 9/x-2. In all my teacher's examples, you only had to multiply the numerator by a number to get the dx, can you multiply by a variable like that?

 

[Ian McPherson]

oh i see, for that term u = x-2, and du = 1. So you would multiply the numerator 9 by (1/9) to make du 1, and multiply the whole term by nine. In the end, the problem would come out to (3/2)x² + 4x + 9ln|x-2| + c right

 

[lookagain]

oh i see, for that term u = x-2,

and du = 1. \(\displaystyle . . . . (du \ = \ 1dx, \ or, \ just \ dx.)\)

So you would multiply the numerator 9 by (1/9) to make du 1, \(\displaystyle . . . . \ (to \ make \ 1dx)\)

and multiply the whole term by nine.

In the end, the problem would come out to (3/2)x² + 4x + 9ln|x-2| + \(\displaystyle > > \)c \(\displaystyle < < \)right

\(\displaystyle (3/2)x^2 + 4x + 9\ln|x - 2| + C \ \ \ \)


\(\displaystyle \text{ \ \ \ \ \ It is standard to capitalize}\)

\(\displaystyle \text{the letter used for this arbitrary constant.}\)