The first thing to note is that the variable v does not appear in either of the integrands, so integration is simple---it's just the difference between the limits.
t=k1Vb-k2(Vf-Vb), where k1 is the first fraction (containing power P as a constant) and k2 is the second fraction (also containing P as a constant).
We should be able to calculate the quantity in the large fractions (effectively k1 and k2) by substituting the given values (apart from P which is to be found).
Start with the numerators: let A=Gvδa/g=1125×1.088/9.8=124.898 approx.
I assume that P in the denominator is the required pressure to be found.
Gvμf=1125 because μf is given as 1.
Let B=½ρCdAc=½×1.2×0.36×2.42=0.52272.
Vb2=218.4484; Vf2=767.29.
The constants can be taken outside the integrals:
7.3=A/(P/14.78-1125-218.4484B)∫[0,14.78]dv-
A/(P/27.7-1125-767.29B)∫[14.78,27.7]dv.
(The limits are parenthesised in square brackets [low,high].)
After integration:
7.3/A=1/(P/14.78-1125-218.4484B)(14.78)-1/(P/27.7-1125-767.29B)(27.7-14.78),
7.3/A=14.78/(P/14.78-1125-218.4484B)-12.92/(P/27.7-1125-767.29B).
Let C=1125+218.4484B=1239.19 approx and D=1125+767.29B=1526.08 approx.
Let a=7.3/A=14.78/(P/14.78-C)-12.92/(P/27.7-D),
a=218.4484/(P-14.78C)-357.884/(P-27.7D),
a(P-14.78C)(P-27.7D)=218.4484(P-27.7D)-357.884(P-14.78C),
aP2-aP(14.78C+27.7D)+409.406aCD=-139.4356P-6051.02068D+5289.52552C,
aP2-P(14.78aC+27.7aD-139.4356)+409.406aCD+6051.02068D-5289.52552C=0.
In the quadratic aP2+bP+c=0 (we can expect up to two solutions):
a=0.05845 approx;
b=-(14.78aC+27.7aD-139.4356)=-3401.77 approx;
c=409.406aCD+6051.02068D-5289.52552C=47931365 approx, and:
P=(-b±√(b2-4ac)/(2a).
P=34,277 or 23,925 approx (calculations should be checked).