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Before you evaluate from x^2 to 0 integrate the function. Since the function you are integrating is a derivative the integral will cancel out the derivative and you will end up just with the expression itself. So:

F(x) = Integrate( Sqrt[t^4+x^3] ) dt = sqrt[t^4 + x^3]

Then evaluate at the interval it asks you to evaluate. However, because you integrated the derivative of a function with respect to t you will need to sub in the interval [x^2,0] for t. So I think the final answer is:

F(x) = sqrt[x^8 + x^3] - sqrt[1 + x^3]

the signs could be backwards depending on the order of the interval. I should like to ask someone else to evaluate my solution.

I misunderstood what you were asking then, I thought you were integrating a differential, but you're just integrating. You should change your topic name.

ok F(x) = Integrate[x^2,0] ( Sqrt[t^4+x^3] ) dt
so I suggest, you start with just integrating an indefinite integral and worry about evaluating it later at which point we can simplify the expression to:
F(x) = Integrate ( Sqrt[t^4+x^3] ) dt

since we are integrating with respect to t, x^3 is a constant, since t has very nicely an exponent that is a multiple of two I'd set u^2 = t^4 and a constant a^2 = x^3 such that your expression goes to:
F(x) = Integrate ( Sqrt[u^2+a^2] ) du/4u^3/2

since du/dt = 4t^3 then dt = du/4t^3 where since u^1/2 = t
dt = du/4u^3/2

Now you should probably integrate by parts. And actually the answer to the book doesn't finish the integral fully it actually leaves it in the integration by parts form. So it should be a simple matter of plugging the values in as per the formula. The formula being:

integrate[f(x)g'(x)]dx = f(x)g(x) - integrate[ f'(x)g(x)] dx

the form of your equation can be just restated as follows:

integrate[ sqrt[u^2 +a^2] * (4u^-3/2) ] du
where f(x) = sqrt[u^2 + a^2] and g'(x) = (4u^-3/2)

by differentiating f(x) and integrating g'(x) we obtain:
f'(x) = 2u/ sqrt( u^2 + a^2) and
g(x) = 8/(3* sqrt)

keep in mind that we still haven't evaluated the integral with the interval given, that is from 0 to x^2 so this is a rough statement, but if we plug these expressions back into the formula for integration by parts we obtain
= sqrt[u^2 + a^2] * 8/(3* sqrt) - integrate[2u/ sqrt( u^2 + a^2) * 8/(3* sqrt)] du

now you can plug in the substitues which we used in the beginning which where to recap:
u^2 = t^4 and a^2 = x^3 and du = dt*4u^3/2 and we also need to evaluate in the interval from 0 to x^2 ok so:

= sqrt[t^4 + x^3] * 8/(3*t^2]) |{0,x^2} - integrate[2t^4 / sqrt(t^4 + x^3) * 8/(3*t^2]) ] dt

note some square roots go away because sqrt(u) = t^2

and... I'm writing out the solution, and it's so close but I made some mistakes along the way very much due to the fact that I was attempting it on the computer. Hwr, I am fairly certain this is the right procedure. I am also, studying for my quantum course, so I will get back to you on this tomorrow when I take another look at the question. Follow the steps and you can probably come to the right solution.

Thanks a lot guys. These posts cleared up some knots. *bow*