What is the definite integral of e^(3lnx)? Answer is (x^4)/4 please show steps....?

e^[3ln(x)]

= e^ln(x³)

= x³, for x > 0

So integrate x³, still subject to the same constraint, x > 0. The given answer neglected that part.

Followup:

I just noticed two other problems. You asked for the definite integral, but that is not possible, having been given no limits of integration. The indefinite integral would have to allow for the addition of any real constant term, but the given answer does not.

∫ e^[3ln(x)] dx = (x^4)/4 + C, x > 0, where C is any real number

I think the question is trying to trick you a bit. If you know logarithmic and exponential properties, then you can easily solve the problem.

a*ln x = ln (x^a)

e^(ln b) = b

Therefore,

y = e^(3*ln x)

y = e^(ln x^3)

y = x^3

Now, you can easily take the integral of a polynomial! :-) Hope you know your integral of a polynomial.

y = x^n

Y = [1/(n+1)] * x^(n+1)

Therefore,

y = x^3

Y = (1/4) * x^4

Y = x^4 / 4

Hint: e^(3lnx) = x^3