How do you integrate dt p2 XYp32 dp lead to t C log 1YXp32

How do you integrate: dt = p^2 /(X-Y*p^3/2) dp lead to:

t= C log (1-(Y/X)*p^3/2) - 2*p^3/2/(3Y)

Solution

We\'ll solve this integral by substitution technique:

We\'ll note X-Y*p^3/2 by t.

X-Y*p^3/2 = t

We\'ll differentiate the left side with respect to p, considering x and y as being constants.

(-3y*p^2/2)dp = dt

Now, we\'ll multiply both, numerator and denominator, by -3y/2 and we\'ll re-write the integral:

(-2/3y)*Int (-3y*p^2/2)dp/(x-y*p^3/2) = (-2/3y)*Int dt/t

(-2/3y)*Int dt/t = (-2/3y)*ln |t| + C

(-2/3y)*Int (-3y*p^2/2)dp/(x-y*p^3/2) = (-2/3y)*ln |x-y*p^3/2| + C

The indefinite integral of the given function is:

Int p^2*dp/(x-y*p^3/2) = (-2/3y)*ln |x-y*p^3/2| + C