Question

How do you find the antiderivative of `int 1/(2x^2-x-3) dx`?

Answer 1

`int 1/(2x^2-x-3) dx = 1/7 ln| (x-2)/(x+3/2) | + C`

Explanation:

First we complete the square of the denominator:
Let `I = int 1/(2x^2-x-3) dx`

`:. I = int 1/(2{x^2-1/2x-3/2}) dx`
`:. I = 1/2 int 1/({(x-1/4)^2 - (1/4)^2-3}) dx`
`:. I = 1/2 int 1/({(x-1/4)^2 - 1/16 - 3}) dx`
`:. I = 1/2 int 1/({(x-1/4)^2 - 49/16}) dx`
`:. I = 1/2 int 1/({(x-1/4)^2 - (7/4)^2}) dx`

Let `u = x-1/4 => (du)/dx=1`, Applying this substitution we get:
`:. I = 1/2 int 1/(u^2 - (7/4)^2) du`

Now a standard integral (that should be learnt if you are studying College Maths) is:
`int 1/(x^2-a^2)dx = 1/(2a)ln| (x-a)/(x+a) |`

And so,
`:. I = 1/2 1/(2(7/4))ln| (u-7/4)/(u+7/4) |`
`:. I = 1/7 ln| ((x-1/4)-7/4)/((x-1/4)+7/4) |`
`:. I = 1/7 ln| (x-2)/(x+3/2) |`