Question

How do you find the indefinite integral of `int (3x^2+4x+1)/(2x)dx`?

Answer 1

`=(3x^2)/4+2x+1/2lnx+C`

Explanation:

Firstly divide each term in the numerator by the denominator.

`int(3x^2+4x+1)/(2x)dx=int[(3x^2)/(2x)+(4x)/(2x)+1/(2x)]dx`

`=int[((3x)/2+2+1/(2x)]dx`

then integrate each term separately

`=(3x^2)/4+2x+1/2lnx+C`