I'd really like some help in solving an integral.
ʃ 1/ √x (1-3√x)
In numerator: 1
In denominator: the square root of x times 1 minus 3 times the square root of x
The answer given is -2/3 ln l1-3√xl + C but I don't know how to get there.
u=1-3√x
du = -3/(2√x) = (-3/2)(1/√x)
you already have 1/√x, so the new integrand becomes
ʃ 1/u (-2/3)du
that should make it clear