Question

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

`int dx/(ax^2+bx+c)` =`1/a int dx/(t^2 +- k^2)`
Please give a brief description.
https://byjus.com/maths/integrals-particular-function/

Number 7 point , how `c/a -(b^2/4a^2)=+-k`

Answer 1

Please see below.

Explanation:

Here ,

`I=int1/(ax^2+bx+c)dx`

`=>I=1/aint1/(x^2+b/ax+ c/a)dx...to(1)`

Now , `x^2+b/ax+c/a` is not a perfect square.

To complete perfect square let `M` be the third term,
such that `x^2+b/ax +M` is a perfect square.

So,

`color(blue)((i)1^(st) term=x^2`
`color(blue)((ii)2^(nd)term=b/ax`
`color(blue)((iii)3^(rd)term=M`

Formula :
`color(blue)(3^(rd)term=(2^(nd)term)^2/(4xx 1^(st) term)`

`M=(b/ax)^2/(4xxx^2)=(b^2x^2)/(4xxa^2x^2)=b^2/(4a^2)`

`i.e. x^2+b/ax+b^2/(4a^2)` is a perfect square.

So,
`x^2+b/ax +c/a=x^2+b/axcolor(red)(+b^2/(4a^2))+c/acolor(red)(-b^2/(4a^2)`

`x^2+b/ax +c/a=(x+b/(2a))^2+(c/a-b^2/(4a^2))...to(2)`

Subst. value of `(2)` into `(1)`

`I=1/aint1/((x+b/(2a))^2+(c/a-b^2/(4a^2)))dx`

Taking , `(x+b/(2a))^2=t^2 and c/a-b^2/(4a^2)=k^2`

`I=1/aint1/(t^2+k^2)dx`,

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

`i.e. sqrt(c/a-b^2/(4a^2))=+-k`

Please note that , according to your question : Number 7 point:

`c/a-b^2/(4a^2)=color(red)(k^2) and` not `color(red)( +-k`