Thank you sir. One more doubt
How to integrate 1/ax^2 +bx

I assume you mean the entire denominator is (ax² + bx), so that the fraction of interest is 1 / (ax² + bx) instead of 1 / (ax²), then plus bx.

Now this is going to be slightly tricky, as we can't simply put the fraction as (ax² + bx)^-1 and integrate.

What we can do, however, is to factorise the denominator into x (ax + b), so that we get

1 / [x (ax + b)]

before we attempt to split them into their respective partial fractions.

Let 1 / [x (ax + b)] = P / x plus Q / (ax + b)

Multiplying throughout by x (ax + b),

1 = P (ax + b) + Qx

Comparing constants,
1 = Pb
P = 1/b

Comparing terms in x,
0x = Pax + Qx
0 = Pa + Q
Q = - Pa
Q = -a/b

So, our fraction becomes

[1 / bx] - a / [b (ax + b)]

which can be written as

(1/b) / x - (a/b) / (ax + b)

Integrating this gets us

(1/b) ln x - (a/b) [ln (ax + b)] divided by a + const
= (1/b) ln x - (1/b) ln (ax + b) + const
= (1/b) [ln x - ln (ax + b] + const

Thank you sir for your explanation