How can you convert this #int dx/(ax^2+bx+c)# into the form of #1/a int dx/(t^2 +- k^2)# ?

Here ,

So,

Where, #color(red)(k^2=c/a-b^2/(4a^2)=>k=+-sqrt(c/a- b^2/(4a^2))#

Kindly take note that, in response to your inquiry: Point number seven:

To convert the integral ( \int \frac{{dx}}{{ax^2 + bx + c}} ) into the form ( \frac{1}{a} \int \frac{{dx}}{{t^2 \pm k^2}} ), where ( t ) is a linear term and ( k ) is a constant, you can complete the square in the denominator of the original integral. Here's how:

1. Write the denominator ( ax^2 + bx + c ) in the form ( a(x^2 + \frac{b}{a}x + \frac{c}{a}) ).
2. Complete the square for the quadratic term inside the parentheses. To do this, add and subtract ((\frac{b}{2a})^2).
3. Rewrite the denominator as a perfect square trinomial plus a constant: ( a\left((x + \frac{b}{2a})^2 - (\frac{b}{2a})^2 + \frac{c}{a}\right) ).
4. Simplify the expression inside the parentheses.
5. Now, the denominator should be in the form ( t^2 \pm k^2 ). You can express it as ( t^2 - k^2 ) or ( t^2 + k^2 ) depending on the sign of the constant term obtained after completing the square.
6. Rewrite the integral using ( t ) as the linear term and ( k ) as the constant term.
7. Adjust the integral by multiplying by ( \frac{1}{a} ) to maintain equality.

This process allows you to rewrite the integral in a form that is easier to integrate, typically through trigonometric substitution or partial fractions, depending on the form of ( t^2 \pm k^2 ).