How can you convert this int dx/(ax^2+bx+c)dxax2+bx+c into the form of 1/a int dx/(t^2 +- k^2)1adxt2±k2 ?

int dx/(ax^2+bx+c)dxax2+bx+c =1/a int dx/(t^2 +- k^2)1adxt2±k2
Please give a brief description.
https://byjus.com/maths/integrals-particular-function/

Number 7 point , how c/a -(b^2/4a^2)=+-kca(b24a2)=±k

1 Answer
Jul 24, 2018

Please see below.

Explanation:

Here ,

I=int1/(ax^2+bx+c)dxI=1ax2+bx+cdx

=>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